**    ;:Mfc 


I 


UNIVERSnYjrCALIFORNIA 
COLLEGE  of  MINING 

DEPARTMENTAL 
LIBRARY 


BEQUEST  OF 


SAMUELBENEDICTCHRISTY 

PROFESSOR  OF 
MINING  AND  METALLURGY 

1885-1914 


*  * 


*.  , 

*       '  *         * 


*    -    * 


* 


The   Compression  and   Transmission 
of  Illuminating  Gas 

A   Thesis   Read    at  the    July,    1905,    Meeting    of    the    Pacific    Coast    Gas 

Association. 


Some  Economics   in   High   Pressure 
Gas  Transmission 

A  Thesis  Read  at  the    September,   1906,  Meeting  of  the  Pacific  Coast  Gas 
Association. 


By  Edward  A.  Rix 

Mem.  Am.  Soc.  M.  E.      Mem.  Am.  Soc.  C.  E.      Assoc.  Mem.   Am.  Soc.    Mining  Eng. 
Mem.  Pacific  Coast  Gas  Assn. 


PUBLISHED  BY  THE 
PACIFIC  COAST  GAS  ASSOCIATION 

SAN  FRANCISCO 
1907 


P  KESS  OF  THE  CALKINS  PUBLISHING  HOUSE 


•  'f 


THE    COMPRESSION    AND    TRANSMISSION    OF 
ILLUMINATING  GAS. 

The  subject  of  illuminating  gas  compression  is  almost 
a  new  one,  and  the  nature  of  the  gas  is  so  entirely 
different  from  that  of  air  that  we  are  obliged  to  con- 
sider the  question  mainly  from  the  theoretical  stand- 
point, backed  up  by  a  few  indicator  cards,  which  have 
been  furnished  us  by  gas  compressors.  But  you  may 
be  assured  that  all  of  the  data  given  herewith  is  emi- 
nently practical,  because  there  has  been  eliminated  all 
of  the  small  variables  that  are  important  from  a  chem- 
ical standpoint,  but  which  the  advancing  piston  of  a 
compressor  cylinder  takes  little  heed  of. 

We  are  riot  concerned  about  the  candle  power  or  the 
commercial  utility  of  a  gas,  but  simply  with  its  weight 
and  composition,  and  what  may  happen  to  it  after  it 
leaves  the  compressor  cylinder  is  not  the  province  of 
this  paper. 

All  gases  are  sponge-like  in  that  they  hold  various 
vapors  from  water  vapor  to  carbon  vapors,  which  they 
lose  to  a  more  or  less  extent  when  the  sponge  is 
squeezed  as  in  the  act  of  compressing  in  a  cylinder, 
and  what  is  squeezed  out  and  how  much  of  it  is  not 
essential  to  our  discussion,  and  lies  better  in  the  realm 
of  the  technical  gas  engineer. 

We  have  assumed,  however,  that  inasmuch  as  when 
we  compress  a  gas  the  temperature  rises  in  a  fixed 
ratio  to  the  pressures,  that  there  is  no  direct  tendency 
for  a  gas  to  change  its  physical  condition  in  the  com- 
pressing cylinder,  for  an  added  temperature  gives  an 
added  capacity  for  saturation,  and  this  probably  in- 
creases in  about  the  same  ratio  as  the  volume  dimin- 
ishes during  compression.  So  that  for  commercial  pur- 
poses we  can  not  be  far  wrong  in  assuming  the  physical 
condition  of  the  gases  as  constant  during  the  range  of 
pressures  that  will  be  ordinarily  met. 

All  phenomena  of  compression    and    expansion    of 


303666 


THE    COMPRESSION   AND    TRANSMISSION    OF    GAS 

gases  is  intimately  associated  with  temperature;  in 
fact,  the  power  to  compress  any  gas  adiabatically  in  foot- 
pounds is  simply  the  difference  in  temperature  between 
the  gas  before  and  after  compression,  multiplied  by  its 
weight  in  pounds,  by  its  specific  heat,  and  then  by  Joules ' 
equivalent  to  convert  heat  units  to  foot-pounds.  Ex- 
pressed algebraically,  this  equation  is: 

L  =  J  W  Cp  (T  —  T0)  where 

J  is  Joules'  equivalent  =  772. 

W  —  the  weight  in  pounds  avoirdupois  to  be  com- 
pressed. 

(7p  is  the  specific  heat  of  the  gas  at  constant  pressure. 

T0  is  the  initial  absolute  temperature. 

T  is  the  final  absolute  temperature. 

L  is  the  work  expressed  in  foot-pounds. 

This  is  the  general  equation  for  the  compression  of 
any  gas  adiabatically. 

In  glancing  at  this  equation,  the  first  stumbling 
block  we  strike  is  CP ,  the  specific  heat  of  the  gas  at  con- 
stant pressure,  and  this  must  be  first  determined. 
After  that  we  must  discover  some  means  of  finding  T, 
the  final  temperature. 

To  anticipate  a  little,  it  may  be  stated  here  that 
these  temperatures  are  all  functions  of  the  ratio  of  the 
specific  heats  of  gas  at  constant  pressure,  and  at  con- 
stant volume. 

It  is  then  our  first  duty  to  understand  about  these 
two  specific  heats  and  to  know  how  to  determine  them 
for  any  gas,  and  the  rest  is  simple. 
r-  The  specific  heat  of  any  substance  is  the  amount  of 
heat  one  pound  of  that  substance  will  absorb  to  raise 
its  temperature  1°  Fah.,  the  specific  heat  of  water 
being  1. 

When  a  gas  is  heated  two  different  results  may  be 
obtained,  depending  upon  whether  the  gas  is  allowed 
to  expand  and  increase  its  volume  when  heated,  the 
pressure  remaining  constant,  or  whether  the  air  is  con- 
fined, the  volume  remaining  constant,  and  the  pressure 
increased.  The  amount  of  heat  to  raise  the  tempera- 
ture of  a  gas  1°  under  these  two  conditions  is  different. 


THE   COMPRESSION   AND   TRANSMISSION   OF   GAS 


7ABLE  J. 


JULY  /905 


/sor/Ytft/viL  Cu/fy£  ft,  vo  •  ft 


0  THE   COMPRESSION   AND   TRANSMISSION   OF   GAS 

therefore  the  specific  heat  is  different.  The  former  is 
called  —  Specific  heat  at  constant  pressure,  and  the  lat- 
ter —  Specific  heat  at  constant  volume. 

Referring  to  Table  1,  Figure  3,  if  we  have  a  cylinder 
A,  containing  one  pound  of  gas  at  atmospheric  pres- 
sure, and  a  piston  P,  without  weight,  but  having  an 
area  of  one  square  foot,  and  heat  the  gas  until  the  tem- 
perature has  risen  1°  Fah.,  the  gas  will  have  expanded 
by  the  small  amount  d  as  in  Figure  4,  and  raised  the 
piston.  This  expansion  is  1/460  of  the  original  volume, 
at  0°  Fah. 

It  is  evident  that  inasmuch  as  the  piston  has  raised 
and  displaced  the  atmosphere,  that  work  has  been 
done,  which  must  have  absorbed  heat  in  addition  to 
that  necessary  to  raise  the  temperature  of  the  air  1°. 
If  the  piston  was  fastened,  as  in  Figure  3,  the  gas 
would  have  required  just  that  less  heat  to  raise  it  1° 
as  was  required  to  lift  the  piston  through  the  distance 
<fc  1/460  of  its  volume.  The  amount  of  heat  required 
in  the  first  instance  is  called  specific  heat  at  constant 
pressure,  and  the  latter  at  constant  volume. 

Specific  heat  of  most  of  the  gases  at  constant  pres- 
sure has  been  determined  by  Regnault  and  others  ex- 
perimentally, and  the  symbol  is  Cp. 

The  amount  of  work  done  in  lifting  the  piston 
through  the  distance  d  is  measured  the  same  as  the 
work  done  by  any  piston  by  multiplying  the  pressure 
on  the  piston  by  the  distance  passed  through.  The 
area  multiplied  by  the  distance  is  the  volume,  which 
may  be  expressed  by  V.  The  distance  d  is  1/460  at  0° 

Fah.^  or  may  be  expressed  by     -—  . 

Let  P  be  the  pressure,  and  R  the  foot-pounds  of 
work  done,  then 

j/  p 

=R  and  this  is  called  the  Simple  Gas  Equa- 


tion, and  about  it  hangs  many  important  deductions. 

R  is  a  constant  for  any  gas,  because  inasmuch  as 
gas  expands  uniformly  for  each  1°  of  heat,  any  volume 
as  Vl  multiplied  by  its  corresponding  Pt  and  divided 


THE   COMPRESSION   AND   TRANSMISSION   OF   GAS 


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8  THE   COMPRESSION   AND   TRANSMISSION   OF   GAS 

by  its  corresponding  temperature  Tl  will  equal  R,  or 
to  put  it  algebraically, 

V.P     yP      V"  P" 


R  being  always  in  foot-pounds,  if  we  divide  it  by 
Joules'  equivalent  772,  which  is,  as  you  know,  the 
amount  of  foot-pounds  equal  to  1  heat  unit,  and  which 
is  always  denoted  by  J,  we  shall  have  the  amount  of 
heat  units  that  were  converted  into  work  to  raise  the 
piston,  and  this  amount  of  heat,  we  know,  must  be  the 
difference  between  the  specific  heat  at  the  constant 
pressure  and  the  specific  heat  at  constant  volume,  or, 

R  _  r      r 

—  —  Cp  —  Cv 

4 

from  which  we  have 
r-r      R 

Cv  —  Cp  --  J 

an  equation  from  which  the  specific  heat  at  constant 
volume  may  be  determined  for  any  gas  within  the 
limits  of  its  stability,  and  certainly  within  the  commer- 
cial pressures  you  are  likely  to  encounter. 

For  a  perfect  gas,  these  specific  heats  are  practically 
constant;  that  is,  they  are  not  affected  by  pressure  or 
temperature,  but  so  far  hydrogen  and  air  appear  to  be 
nearer  than  any  other  gases.  CO  and  C02,  which  are 
inferior  components  of  illuminating  gas,  as  it  is  now 
made,  show  the  greatest  deviation,  but  not  enough  to 
render  their  vagaries  of  moment  in  the  consideration  of 
the  power  question,  consequently  all  the  following  data 
have  been  calculated  on  the  basis  of  the  simple  gas  law. 

PV 

~—  —R  —Constant. 

As  an  example  showing  how  to  calculate  the  specific 
heat  at  constant  volume,  let  us  take  C2  H4.  This  gas 
have  been  calculated  on  the  basis  of  the  simple  gas  law. 
values  ascribed  to  Regnault  in  the  references  we  have 
at  hand. 

Upon  apptying  the  simple  gas  equation  to  the  Reg- 


THE    COMPRESSION   AND    TRANSMISSION   OF    3AS  9 


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10          THE    COMPRESSION    AND    TRANSMISSION    OF    GAS 

nault  value  there  was  a  large  discrepancy,  and  it  will 
be  interesting  no  doubt  to  make  the  calculations  here, 
and  thus  make  them  serve  the  double  purpose  of  show- 
ing how  to  determine  the  specific  heat  at  constant 
volume  and  to  point  out  the  error. 

Regnault  gives  the  Cp  of  C»  H4  to  be  .404,  and  Cv  to 
be  .173.  The  weight  per  cubic  foot  to  be  .0780922,  or 
12.8  cubic  feet  in  one  pound  at  32°  Fah. 

If,  now,  one  pound,  or  12.8  cubic  feet,  be  heated  to 
1°  Fah.  and  allowed  to  expand,  the  simple  gas  equation 

P-^=  R  will  give  at  32° 
14.7  X  144  X  12.8 


=  R  =  55. 


492 

Fifty-five  foot-pounds  of  work  has  been  performed 
by  the  gas  in  expanding  against  the  atmosphere;  to 
convert  this  into  heat  units  we  divide  by  Joules'  equiv- 
alent, 772. 

—=.07124  units  of  heat. 
772 

Inasmuch  as 

D  & 

=  07124 


we  have  Cv  =.404—  .07124=.  3327,  instead  of  .173  as 
determined  by  Regnault.  The  ratio  between  the  two 
specific  heats  forms  the  basis  for  all  the  calculations  for 
the  relations  between  pressure,  volume,  and  tempera- 
ture in  compressing  gas,  and  that  is  why  we  must  be 
particular  about  these  specific  heat  factors. 

—    =  y  (gamma),  which  we  shall  discuss  further 
C\ 

on,  and  which  is  brought  in  now  simply  as  additional 
proof  about  the  figures  which  we  have  just  obtained  for 


For  C2  HV  using  Regnault  's  values,  we  have 
404 
.173 
for  our  values 


=          = 


THE   COMPRESSION   AND   TRANSMISSION   OF   GAS 


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12          THE    COMPRESSION    AND    TRANSMISSION    OF    GAS 


In  reading  a  new  book  by  Travers  on  the  study  of 
gases  (page  275),  he  gives  some  very  interesting  calcu- 

lations to  show  the  limiting  values  of  —^  or  -y, 

C\ 

His  conclusions  are  that  for  a  monoatomic  gas  within 

PV 

the  limits  of  the  simple  gas  equation  —  -=R,  the  val- 

ues of  -  --  can  never  exceed  1.667,  and  the  value  for  a 
C  \- 

diatomic  gas  should  range  about  1.4  and  the  polyatomic 
gases  still  less,  until  we  reach  the  value  of  1,  where,  of 
course,  there  should  be  no  expansion  work  at  all  when 
heat  was  applied. 

We  can  see,  therefore,  that  the  value    —  of  2.33  from 

C\ 

Regnault's  values  is  an  impossibility,  the  maximum 
possible  value  being  only  1.667,  and  C2  H±  being  the 
polyatomic  gas,  its  value  would  be  less  than  1.4,  all  of 

which  indicates  that  our  figures  —  =1.214  are  approxi- 

cV    • 

mately  correct. 

It  will  now  be  necessary  to  apply  our  understanding 
of  these  principles  and  try  and  determine  the  values  of 
the  specific  heats  for  illuminating  gas.  There  seems  to 
be  plenty  of  data  about  the  specific  heat  at  constant 
pressure  for  gas  mixtures,  but  nothing  about  the  spe- 
cific heat  at  constant  volume. 

Reference  is  now  made  to  the  Tables  2,  3,  4,  5,  6,  7, 
and  8,  which  show  the  composition  and  heat  properties 
of  seven  different  g'ases  and  the  methods  employed  in 
determining  the  weights,  specific  gravities,  and  specific 
heats. 

Column  1  is  the  chemical  symbol  for  the  different 
components. 

Column  2  is  the  percentage  by  volume  of  the  differ- 
ent' components. 

Column  3  gives  reliable  weights  per  cubic  foot. 


THE    COMPRESSION   AND    TRANSMISSION    OF    GAS          13 


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THE   COMPRESSION   AND   TRANSMISSION   OF   GAS 


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THE   COMPRESSION   AND   TRANSMISSION   OF   GAS          15 


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THE   COMPRESSION   AND    TRANSMISSION   OF   GAS 


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THE    COMPRESSION   AND    TRANSMISSION    OF    GAS 


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SUAI  CAL  DECZ 


18          THE    COMPRESSION   AND    TRANSMISSION   OF   GAS 

Column  4  gives  the  specific  heat  of  each  component 
gas  as  determined  by  Regnault  and  others. 

Column  5  gives  the  product  of  the  different  percent- 
ages of  the  component  gases  and  their  weights  per 
cubic  foot,  or  Column  2  multiplied  by  Column  3.  The 
total  sum  divided  by  100  gives  the  weight  of  the  gas 
per  cubic  foot. 

Column  6  gives  the  product  of  Column  4  and  Column 
5  for  specific  heat,  being  a  weight  function.  We  must, 
in  order  to  get  the  specific  heat  of  the  compound  gas, 
take  into  consideration  not  only  the  percentages  of  the 
component  parts,  but  the  weights  as  well,  and  also  the 
specific  heat  of  each  component.  The  sum  of  the 
products  in  column  divided  by  100,  and  then  by  the 
weight  of  one  cubic  foot  of  the  compound  gas,  will  give 
the  specific  heat  at  constant  pressure  Cp. 

Column  7  gives  the  calculations  to  find  the  specific 

C 

heat  at  constant  volume  and  also  R  and-^  or  y  for  each 

0  v 

gas,  and  also  various  factors  of  y  which  we  will  find 
useful  later. 

Table  9  concentrates  Tables  2  to  8,  so  that  we  may 
study  them  easier. 

You  will  note  that  our  results  cover  quite  a  field, 
taking  in  California  fuel  oil  gas,  Massachusetts  coal 
gas,  Indiana  natural  gas,  California  natural  gas,  anJ 
California  carburetted  water  gas,  and  after  carefully 
studying  their  heat  and  power  properties,  as  shown  in 
Table  9,  we  have  selected  the  fuel  oil  gas  made  in  Oak- 
land as  having  the  best  average  properties  for  the  pur- 
poses we  have  in  view,  and  particularly  as  fuel  oil  gas 
is  the  one  you  will  probably  have  most  to  deal  with. 

We  may  therefore  consider  our  subject  as  having  for 
a  basis  a  gas  with  the  following  properties  at  32°  Fah. : 
Weight  per  cubic  foot,  .0323577. 
Cubic  foot  in  one  pound  avoirdupois,  30.98. 
Specific  gravity,  .4008. 

Cp  =.6884. 
Cv  =.5159. 


THE   COMPRESSION   AND   TRANSMISSION   OF   GAS          19 


#=133.2. 
£=8467  I  •——: 


A  cubic  foot  of  gas  varies  in  weight  according  to 
the  altitude  or  pressure,  and  also  according  to  the  tem- 
peratures. The  law  of  this  variation  is  expressed  as 
follows  : 

Having  given  the  weight  of  a  gas  for  any  tempera- 
ture, or  any  pressure,  then  the  weight  at  any  other 
temperature  or  pressure  will  be  as  the  ratio  of  absolute 
temperature  or  pressure,  or 

W'=W  -^.  or    W^  where 

W=  known  weight. 

T°  and  P°  the  known  temperature  or  pressure  and 
W  the  desired  weight. 

For  example  :  —  Our  standard  gas  weights  at  sea 
level,  or  14.7  pounds  absolute  pressure,  and  32°  Fah., 
.03235  pounds  per  cubic  foot  ;  at  20  pounds  gauge,  or 
34.17  pounds  absolute,  a  cubic  foot  would  weigh  .03235 

X  ^^  =.03235  X  2.36=.076346  pounds,  and  at  60° 

Fah.,  instead  of  32°  Fah.,  this  cubic  foot  would  weigh 

520 
.076346  X          =.0819  pounds,  460  being  the  absolute 


temperature  of  O°  and  520°  the  absolute  temperature 
of  60°  Fah.=460°  +60°=520°. 

Altitudes  are  nothing  more  or  less  than  pressures 
less  than  sea  level,  and  are  treated  just  the  same  as 
pressures  above  the  normal  atmospheric. 

Thus  at  5225  feet  the  absolute  pressure  is  12.044,  con- 

12  044 
sequently    gas    at    this    altitude    would    weigh         ' 

times  the  weight  at  sea  level. 


20          THE    COMPRESSION   AND    TRANSMISSION    OF    GAS 

For  your  convenience  it  may  be  well  to  add  here  that 
when  the  barometric  pressure  is  known,  the  atmos- 
pheric pressure  is  found  by  multiplying  the  barometric 
pressure  by  .4908,  or  P°  =  B  X  -4908. 

For  example. — When  the  barometer  is  29.92  the 
atmospheric  pressure  is  29.92  X  .4908,  or  14.7,  the  nor- 
mal sea-level  pressure. 

To  find  the  atmospheric  pressure  when  the  altitude 
in  feet  is  given,  we  have 

,  70          57000  N—N* 

p  100,000,000     m  whlch 

N  =  altitude  in  feet. 

For  example. — To  find  the  atmospheric  pressure  at 
10,000  feet  we  have 

P«_1A  79  57,000X10,000— (1Q,OQO)2 

100,000,000 

P°  =  14.72  --  4.7  =  10.02,  the  atmospheric  pres- 
sure required. 

The  foregoing  rules  will  be  all  that  is  necessary  to 
calculate  all  variations  of  weights  due  to  pressure, 
altitude,  or  temperature,  and  relative  volumes  follow 
exactly  the  same  laws  as  relative  weights. 

For  convenience  in  many  calculations  Table   10   is 

p 
given  herewith,  showing  the  pressure  ratios,  or  — -  for 

every  pound  from  1  to  110,  and  the  volumes  ratios  will 
be  inversely  as  the  pressure  ratios  and  consequently 
the  reciprocal  of  the  figures  on  the  table. 

This  might  be  called  a  table  showing  also  the  rates  of 
isothermal  compression  or  expansion  or  Marriotte 's 
law.  the  general  formula  for  which  is : 

P°  V°  -  P  V  -  Constant,  or  in  other  words,  the 
product  of  any  pressure  by  its  volume  is  always  equal 
to  the  product  of  any  other  pressure  by  its  volume,  and 
this  rule  will  be  found  useful  in  determining  the  con- 
tents of  receivers,  etc.  It  must  always  be  remembered 
that  in  using  these  rules  all  temperatures  must  be 
alike,  or  corrections  made  according  to  the  rules  just 
given. 


THE    COMPRESSION   AND    TRANSMISSION    OF    GAS 


21 


TABLE  jo 


Gauge 


Gauge 


fio 


Giagc 


/066O27 


38 


3S3SO2.6 


75 


//36054- 


3  (55305  '3 


76 


3-7Z/OGO 


77 


GJS3G073 


3-7&9407 


78 


3  3*7  S  34- 


79 


6  374/33 


ao 


/4-76/Q9 


3  333/SG 


a/ 


6-5/0/37 


+S 


32 


46 


83 


JO 
/  i 
/2 


34 


7+323  f 


3S 


4-  3333SS 


63S03JZ2 


37 


33 


6  •336376 


4  f  37404 


39 


2O3Q+32 


S3 


<#  60&+3/ 


30 


7/22430 


2/S6+S9 


7/3O4  4T7 


4-  74  J  4SJ- 


93 


£  360&+0 


57 


7  334&3S 


<SO 


97 


7-S936/S 


24 
25 
£6 


G/ 


&-/43G4  7 


93 


2  7  O  06  73 


33 


7-734-073 


2  7686O2 


<53 


7SOZ  70O 


2336729 


6* 


/O' 


737O  7X7 


23 
23 
30 


GS 


&9727G3 


Gff 


3  O-4OS/0 


67 


S3S7&09 


S  074330 


63 


32 


3176364- 


69 


33 
34 
3S 


7O 


3  3J23/G 


72 


£4/4-943 


36 


3448372 


73 


3  4323  7O 


7-f 


6  O33&9& 


22          THE   COMPRESSION   AND   TRANSMISSION   OF   GAS 
ISOTHERMAL   COMPRESSION. 

There  are  two  methods  of  compressing  any  gas. 

First. — Where  the  temperature  remains  unchanged 
during  compression.  This  is  called  isothermal  com- 
pression and  is  the  ideal  method  never  realized  in 
practice. 

Second. — Adiabatic  compression,  which  is  the  kind 
we  meet  in  practice  where  the  heat  developed  by  com- 
pression expands  the  air  being  compressed  until  it  fol- 
lows a  different  law  from  Marriotte. 

While  isothermal  compression  is  not  practical,  it  is 
necessary  to  know  about  it  and  how  to  make  the  calcu- 
lations concerning  it. 

We  have  found  that  the  volume  ratios  are  inversely 
as  the  absolute  pressure  ratios  in  isothermal  compres- 
sion. Consequently  if  the  pressure  ratios  are  1,  2,  3, 
and  4,  the  corresponding  volumes  will  be  1,  %,  %,  14. 
To  show  this  graphically,  refer  to  Table  1,  Figure  1. 

Let  A  B  be  the  line  of  0  pressure  or  the  perfect- 
vacuum  line.  C  D  the  intake  line  and  we  erect  pres- 
sures ordinates  GH  =  2XDBatsL  point  H  equal  to 
!/2,  A  B  and  7J  =  3X##ata  point  /  =  %  A  B  and 
E  K  —  4  X  #  #  at  a  point  K  —  1/4  of  A  B  counting  all 
volumes  from  F  B  or  the  end  of  the  piston  stroke. 

If  we  join  the  points  C  G  I  E  in  a  curved  line,  it  will 
be  the  isothermal  or  logarithmic  curve  and  it  will  be 
noted  that  the  area 

EFBK  =  4:XV4:  =  ^ 
I  L  B  J  =  3  X  %  =  1. 
0JfJBJI  =  2X%  =  l 
C  D  B  A  —  1  X  1  or  1 

As  found  before,  P°  Y°  =  P'  V  =  Constant,  and  the 
figure  represents  the  ideal  indicator  card  for  isother- 
mal compression  for  four  compressions,  counting  from 
0,  and  the  above  method  will  always  be  proper  to  lay- 
out an  isothermal  curve,  no  matter  what  the  intake 
pressure  may  be. 

To  find  the  work  of  compression  and  delivery  iso- 
thermally 


THE   COMPRESSION   AND   TRANSMISSION   OF   GAS          23 


P 

P°  V°  hyp.  log.  —  in  foot-pounds  in  which 

P°  =  Initial  pressure  absolute. 
y°  =  Initial  volume. 
P  —  Final  pressure. 
L  =  Work  required. 

In  all  our  calculations  V°  will  be  taken  as  one  cubic 
foot. 

For  example: — How  many  foot-pounds  of  work  are 
required  to  compress  1  cubic  foot  of  gas  at  sea  level  to 
eighty  pounds  gauge  pressure? 

For  sea  level  P°  per  square  foot  =  14.7  X  144  — 
2116.8  pounds.  Then 

/, =2116.8  hyp.  log.    -~o 

Consulting  Table  10,  we  find 

p 

po  for  80  pounds  gauge=6.442  the  hyperbolic  loga- 
rithm of  which  is  1.863. 

Substituting,  we  have 

L  ==  2116.8  X  1.863  =  3943  foot-pounds. 

If  a  table  of  hyperbolic  logarithms  is  not  at  hand,  it 
would  be  well  to  remember  that  hyp.  log.  =  common 
log.  X  2.3026. 

3945 
The  H  P  required  for  above  work  will  be  — — —  == 

.1195  H  P. 

To  find  the  M  E  P  of  isothermal  compression, 

p 
M  E  P  =  P°  hyp.  log.    -—    using  the  quantities  in 

the  previous  example  we  have  M  E  P  =  14.7  X  1.863 
=  27.38  pounds.    We  know  that  HP  =    M  f JT  *  - 

ooUUU 

and  for  one  cubic  foot  V  =  1  X  144.     Consequently, 
using  the  last  example, 

27.38  XIX  144 
//  P  —         "QQnTvT"          ~  .1195,  the  same  result 


24          THE   COMPRESSION   AND   TRANSMISSION   OF   GAS 

1  X  144 
as  before.        QQQQQ     =  .00436.     Consequently  a  short 

and  convenient  formula  would  be  for  isothermal  com- 
pression H  P  =  .00436  X  M  E  P. 

It  will  be  noted  that  none  of  the  physical  properties 
of  gases  enter  into  the  above  equations,  consequently 
we  must  conclude  that  it  takes  the  same  power  to  com- 
press one  cubic  foot  of  any  gas  isothermally  to  the 
same  pressure,  provided  the  ratios  of  pressures  are  the 
same. 

ADIABATIC   COMPRESSION. 

We  have  before  stated  that  isothermal  compression 
is  ideal,  and  not  realized  in  practice.  All  of  the  work 
expended  in  compressing  a  gas  is  converted  into  heat 
instantly,  and  this  increases  both  the  temperature  and 
the  volume  of  the  gas  during  compression,  so  that,  in- 
stead of  having  a  relation  between  pressure  and  vol- 
ume (P0  V0  =  P  V  =  Constant),  such  as  we  found  in 
isothermal  compression,  we  now  have  a  relation  P0  V0y 
—  P  Vy  =  Constant,  or  in  other  words,  the  gamma 

powers  of  each  volume,  multiplied  by  its  corresponding 
pressure,  is  Constant.  This  is  the  equation  of  the 
Adiabatic  curve,  y  is  the  same  that  we  found  to  be 
the  ratio  between  the  specific  heat  at  constant  pressure 
and  that  at  constant  volume.  This  relation  can  per- 
haps be  fastened  a  little  easier  in  the  mind  by  remem- 
bering that  the  equation  of  the  isothermal  curve  rep- 
resents the  law  of  Marriotte  and  the  equation  of  the 
adiabatic  curve  represents  the  Exponential  law  of  Mar- 
riotte. 

Inasmuch  as  the  power  to  compress  a  gas  is  meas- 
ured practically  by  the  indicator  diagram,  and  this 
in  turn  is  compared  to  the  adiabatic  curve  which  is 
theoretical  curve  of  compression,  and  inasmuch  as  we 
depend  upon  the  value  of  y  to  construct  this  curve,  it 
will  be  at  once  seen  why  we  were  so  particular  to  dis- 
cover the  relation  ~  =  r.  Now  if  Pn  VJ  =  P  Vy  and 


THE    COMPRESSION   AND    TRANSMISSION    OF    GAS          2o 


V°  P°       V  P 

irom  the  simple   gas  equation     __-_  _  ^—     =  E,  by 

combining  these  we  have  all  the  adiabatic  relations  be- 
tween volume  pressure  and  temperature  as  follows: 


p_     /FV_  (L\y -y, 

P°       \V )      ~  \ToJ 


y-i        /  P  \  y^i 

y 

It  will  always  be  necessary  to  use  the  above  formu- 
las in  making  calculations  for  pressures,  temperatures, 
and  volumes,  or  for  power  to  compress  any  gas  which 
varies  far  enough  from  the  standard  we  have  selected 
to  make  it  necessary,  but  there  is  no  doubt  that  for  all 
practical  purposes,  at  least  for  the  present,  Table  11, 
which  is  calculated  for  our  standard  gas,  will  give  the 
proper  values  for  rapidly  and  easily  calculating  any 
problems  connected  with  compressing  illuminating  gas. 

All  reference  to  expansion  is  purposely  omitted,  be- 
cause gas  will  probably  never  be  used  for  expansion 
work  in  an  engine  as  air  is  used. 

Assuming  that  all  may  not  be  familiar  with  just  how 
to  arrive  at  the  results  as  indicated  in  Table  11,  let  us 

p 

take  a  ratio  of  —Q  —  2  corresponding  to  14.7  pounds 

V 

gauge  pressure  and  discover  what  are  the  values  of  -y-0 

T  V  /  P°  \  i 

and   — .    We  have  —  I  —  I  -        '  y  we  have  already  de~ 
T°  F°  \  P  /  y 

cided  from  our  standard  gas  to  be  1.334. 
Therefore,      y  =  ]~^  =  .749        ~Q=  (^  '? 


since 


po 

— —  —i/o  or  .5  we  have 


26          THE   COMPRESSION   AND   TRANSMISSION   OP   GAS 
y  .749 

—  =  .5  or 

y 

Log.—  ==log  .5  x.749. 

Log.  .5  =  1.6989  X  .749  =  1.77447  =  log.—  giving 

y  yo 

value  of  -=^  =  .5949  and     =-     will  be  reciprocal  of 

~   or  1.681. 

To  find  the  ratio  of  temperature  for  this  same  rate 

y-l 

of  compression,  we  have  -^=  •  —  I   y       ~— -  =  .25. 

r°    \poj        y 

Hence  : 
Log.-|ro=.251og.  ~ 

Log.  ^  =  .301   X    -25  =  .07525  =  Log.  ~ 

T 

—  1892 

— .±OV6. 

T  =  520  X  1.1892  =  618°  absolute  or  158°  Fah.  if 
T°  =  60°  Fah.    We  have  then 
p  yo  y 

O  1    £Q~\  F 

-=—   ~          -^T    J..DOJ.  TV~  ••• 


=  1.1892      T  =  158° 

Air  under  the  same  conditions  gives 

P  V°  V 

-  =  2      {,=1.6349      ~0  =  . 

^  =  1.2226      T  =  175°  Fah. 

These  examples  will  serve  to  show  how  this  Table  11 
was  calculated.  A  few  examples  will  show  its  use. 

Problem. — To  find  the  final  temperature  due  to 
adiabatic  compression. 

P  T 

Opposite   —   and    under    the    headline  — -    will  be 

found  the  ratio  of  absolute  temperatures. 

Example. — What  is  the  final  temperature  due  to  14.7 
pounds  gauge  pressure  at  sea  level  and  60°  Fah.  ? 


THE    COMPRESSION   AND    TRANSMISSION    OF    GAS          27 


7*  OLE  //.               /lOIAMT/C  TA8L  £  fO/f  G*S                  J<JL  Y  /$  Ot> 

/> 
/>* 

T 
To 

T     , 

ro 

-L. 

Vo 

JVUf1B£# 

Dirr. 

JVi/HO£/r 

Oiff. 

/* 

/0+66 

-+/2 

046G 

BO  63 

30O 

/4 

/037& 

369 

0373 

7763 

733 

/-6 

//£+7 

336 

/£+7 

703O 

J3+ 

/3 

//fa  3 

JO9 

/fS3 

•6436 

433 

£ 

/  /39£ 

£37 

sasz 

J346 

•4JO 

22 

/£./73 

260 

•£/79 

JS33 

3f£ 

£+ 

/  £+47 

£?/ 

£447 

•f'S6 

SOI 

£6 

/•2G36 

£63 

£638 

•433S 

£64 

za 

/Z3&* 

S35 

£366 

462/ 

£33 

3 

/3S6/ 

£/* 

3/6/ 

4368 

£07 

3£ 

/337S 

£04 

JJ7S 

4/S/ 

/46 

3  + 

/  '3573 

/as- 

•3373 

3935 

s&a 

36> 

/  3774 

/30 

3774 

•3327 

/jse 

J<3 

/3962 

S<3O 

3S62 

•3675 

/J9 

4- 

/  4-/4-Z 

S73 

4J4£ 

•3536 

/££ 

*£ 

/43/S 

/6S 

•+3/S 

•34/O 

//7 

+  4- 

A++33 

S£Z 

•4433 

•3233 

/OS 

<+6 

/  4645 

/J7 

464S 

3/35 

33 

<+a 

/+6O£ 

/S£ 

•4<90£ 

•3O36 

34 

5 

/*3S+ 

697 

•49f4 

£392 

363 

6 

SS65/ 

6/5- 

S6S/ 

•£603 

£0* 

7 

/•6266 

tt£ 

ff£66 

£32* 

£ZZ 

8 

S6&/0 

f&3 

6d/3 

•£/O3 

/77 

3 

/•732/ 

-462 

•732/ 

•/3£6 

/*7 

/o 

/  7733 

7733 

•/773 

£t\I?'* 

28          THE    COMPRESSION    AND    TRANSMISSION   OF    GAS 

P        29  4  T 

p=^  =  2.  Then  £.  ='1.1892,    or     520  X  1.1892  = 

618°  abs.  or  158°  Fah. 

If  the  initial  temperature  has  been  100°  then  560  X 
1.1892  =  666°  abs.  or  206°  Fah. 

It  is  readily  noted  from  this  that  the  higher  the  ini- 
tial temperature,  the  higher  the  final  temperature,  and 
it  will  also  be  noted  that  while  there  is  a  difference  of 
40°  between  the  initial  temperature,  there  is  a  differ- 
ence of  48°  between  the  final  temperatures ;  a  differ- 
ence of  8°. 

Inasmuch  as  the  temperature  developed  during  com- 
pression is  at  the  expense  of  power,  it  is  evident  that  it 
takes  more  power  to  compress  the  same  weight  of  gas 
at  100°  Fah.  than  at  60°  Fah.  to  the  same  pressure,  all 
other  conditions  being  similar. 

It  is  an  axiom,  therefore,  that  the  cost  of  power  for 
compressing  gas  will  be  the  least  when  the  initial  tem- 
perature is  the  lowest,  and  it  will  be  shown,  later  on, 
that  cooling  before  compression  will  effect  a  consider- 
able saving,  if  the  gas  to  be  compressed  is  drawn  from 
the  holder  exposed  to  the  sun,  provided,  of  course,  that 
cooling  water  may  be  had  at  a  small  expenditure  of 
power. 

Problem. — To  find  the  volume  immediately  after 
compression. 

y 

Consult  Table  11,   and  under  the  heading    — -•    and 

p 

opposite  the  pressure  ratio    —    the  proper  value  will 

be  found ;  and  it  must  always  be  remembered  that  these 
values  of  temperature  and  volumes  assume  no  radia- 
tion of  heat  whatever,  for  when  the  heat  generated  by 
compression  has  radiated  the  temperatures  and  vol- 
umes are  as  calculated  isothermally. 

V 

Please  note  that  ----  is  measured  from  the  end  of  the 

stroke.     The  difference  given  in  Table  11  will  enable 

p 
greater  or  lesser  values  of  —  to  be  conveniently  deter- 


THE    COMPRESSION    AND    TRANSMISSION    OF    GAS          29 

mined  by  simple  rules  of  proportion. 

From  this  table  the  adiabatic  curve  can  be  readily 
drawn. 

Refer  to  Table  1,  Figure  2. 

Let  A  B  be  the  intake  line  and  C  D  the  line  of  0  pres- 
sure, these  lines  representing  the  piston  stroke.  Divide 
A  B  into  a  decimal  scale;  beginning  at  B  erect  F  D  at 
the  end  of  the  stroke  and  divide  it  into  equal  values 
of  B  D.  B  D  may  be  the  value  at  sea  level  or  at  an 
altitude  or  it  may  be  any  intake  pressure  whatever; 
these  rules  will  always  apply.  These  values  of  B  D 
may  be  subdivided  into  five  parts,  where  special  accu- 
racy is  required,  and  their  values  will  also  be  found  in 
Table  11. 

p 

D  H  representing  a  ratio  of  —  =  2,  the  correspond- 
ing value  of  yo  will  be  found  in  Table  11  to  be  .5948, 

and  laying  off  the  value  the  point  S  will  be  found. 

P  V 

Similarly  at  G  representing     —    =  3  we  find    y£= 

.4388,  and   laying  this   off   we   find   that    the  point  M. 

P  v 

And  then  F  representing    -—    =  4  has  a  value  for 

P°  yo 

of  .3536,  and  we  lay  off  this  value  and  find  point  J. 
Joining  the  point  J  M  S  A  we  develop  the  adiabatic 
curve,  and  the  shape  of  this  curve  will  depend  upon 

C 
the  length    of    the  card,  the  value  of    ~     or  v.    The 

Cv 

equation  of  the  curve  is  P  Vy  =  P'  Vn  or  referring  to 
the  diagram. 

M  0  X  (M  G)y  =  JLX  (J  F)y 

Problem. — To  determine  the  power  to  compress  a 
gas  adiabatically. 

All  that  precedes  this  subject  has  been  necessary  to 
its  proper  understanding,  and  while  possibly  the  va- 
rious symbols  are  well  remembered,  it  will  probably 


30          THE   COMPRESSION   AND   TRANSMISSION   OF   GAS 

be  better  to  group  them  together,  so  that  they  may  be 
readily  referred  to. 

P°  is  always  the  lesser  absolute  pressure,  and  conse- 
quently the  intake  pressure  in  compression.  We  shall 
take  this  as  14.7  at  sea  level,  for  the  4-inch  water  pres- 
sure of  the  gas  will  not  fill  the  cylinder  at  any  greater 
than  atmospheric  pressure.  P  is  the  final  absolute 
pressure. 

T°  is  the  initial  absolute  pressure,  and  unless  other- 
wise specified  is  taken  at  60°  Fah.  or  520°  absolute, 
that  temperature  being  the  probable  temperature  of 
the  gas  mains. 

T  is  the  final  absolute  temperature. 

y°  is  the  volume  at  P°. 

V  is  the  volume  at  P. 

p>  y  T'  are  intermediate  pressures,  temperatures, 
and  volumes. 

L  is  the  work  expressed  in  foot-pounds. 

H  P  is  horsepower. 

M  E  P  is  mean  effective  pressure,  which  is  always 
gauge  pressure. 

W  is  the  weight  of  a  unit  volume  or  one  cubic  foot 
of  our  standard  gas  at  60°  Fah.  and  at  sea  level,  with 
an  absolute  pressure  of  14.7  Ibs.  per  square  inch,  or 
2116.8  pounds  per  square  foot,  and  equals  .03063 
pounds  avoirdupois. 

J  is  Joules'  equivalent  taken  at  772  foot-pounds. 

(7P  is  the  specific  heat  at  constant  pressure  =  .6884. 

(7V  is  the  specific  heat  at  constant  volume  =  .5159. 

j/is  ~  =  1.334. 

^-=4.     ^i— .25     i-.7S 

j/-i  y  y 

J  CP  =  772  X  -6844  =  531.45  foot-pounds. 

This  value  is  Joules'  equivalent  for  1  Ib.  of  gas  at 
constant  pressure. 

/  W  CP  =  531.45  X  .03063  =  16.28  foot-pounds  = 
Joules'  equivalent  for  1  cubic  foot  of  gas. 

J  W  CP  T°  —  16.28  X  520  =  8465  foot-pounds. 


THE    COMPRESSION   AND   TRANSMISSION   CF   GAS          31 

JL    x  P°  T70  =  4  X  144   X  1  X  14.7  =  8465 

foot-pounds  =  the  intrinsic  energy  of  1  cubic  foot  of  gas 
at  60°  Fah.  and  atmospheric  pressure  at  sea  level; 
or  to  reduce  those  values  of  foot-pounds  to  horsepower, 
we  have 


y-\  33000 

All  of  these  foregoing  quantities  are  constants  to  be 
used  in  determining  the  power  to  compress  gas,  and  as 
we  have  said  before,  are  all  based  on  a  quantity  of  1 
cubic  foot  of  our  standard  gas  at  sea  level  and  60° 
Fah. 

We  mentioned  at  the  beginning  of  this  paper  that 
the  power  to  compress  any  gas  adiabatically  might  be 
expressed  by  the  general  formula 

L  =  J  W  Cp  ( T—  T°) ,  or  to  put  it  in  another  form, 


Sri) 


L  =  J  W  C*l 

You  now  at  once  recognize  the  prefix  J  W  Cp  T°  as 
the  one  for  which  we  have  found  a  value  of  8465  foot- 
pounds. Therefore,  for  our  standard  gas  we  have 

L  —  8465  |  —  — l    J  which  is  a  practical  formula. 

T 

You  also  recognize  that     —  is  all  you  need  solve,  and 

these  values  are  all  given  in  Table  11  for  the  various 

p 

values  of  — .    We  can  now  understand  our  first  prob- 
lem. 

How  many  foot-pounds  are  necessary  to  compress  1 
cubic  foot  of  our  standard  gas  to  14.7  pounds  gauge 
pressure  ? 

P  2Q4- 

4r  =  ^  r^  =  2-    Consulting    Table  11   we   find 
P°         14.  / 

j^  —1  —  .1892  and  8465  X  -1892  =  1601.57  foot- 
pounds, and  the  same  method  may  be  applied  for  all 
pressures. 


32          THE   COMPRESSION   AND   TRANSMISSION   OF   GAS 

If  we  use  the  value  of  J  W  Cp  T°  in  horsepower,  we 
have  HP—  .2564  |  — 1  )  a  perfectly  practical  for- 
mula for  1  cubic  foot  of  our  standard  gas  at  60°  Fah. 
and  at  sea  level. 

Our  previous  example  would  then  be  rendered: 

L  =  .2564  X  .1892  =  .0485  horsepower  for  1  cubic 
foot  compressed  to  14.7  Ibs.  gauge.  At  80  Ibs.  gauge 
pressure. 

j-  —  6.442  and  yo  =  1.593. 

II  P  =  .2564  X  -593  =  .1520  horsepower  per  cubic 
foot,  or  15.20  H  P  per  100. 

MEAN   EFFECTIVE  PRESSURES. 

It  will  be  found  that  inasmuch  as  we  learn  from  an 
indicator  wThat  our  gas  compressor  is  doing,  and  inas- 
much as  M  E  P  pressures  are  quickly  determined  by  a 
planimeter  from  an  indicator  card,  that  to  become 
familiar  with  what  the  M  E  P  should  be  and  compare 
it  with  what  the  compressor  is  doing  is  the  best  prac- 
tical way  of  dealing  with  the  subject. 

We  found  that 

L  =  J  W  <7p  T°  (L—l  \  and  that , 
J  W  (7p  T°  =  ^-P°  T70,  therefore 


L  must  always  equal  M  E  P  X  V°,  we  have 
ME  PXV°  =  -P° 


ft  \ 

M  E  P  =-2—  P°  I  —-_  1  1  and  since 

y-1        V        / 

2—  =  4,  we  have  for  our  standard  gas 


Take  80  Ibs.  gauge  pressure. 


THE    COMPRESSION    AND    TRANSMISSION    OF    GAS          33 

T 

— —  1  =  .593  as  determined  in  a  former  exam- 
ple by  Table  11. 

P°  =  14.7. 

M  E  P  =  4  X  -593  X  14.7  =  34.86  Ibs.  per  sq.  in. 

For  our  standard  gas  for  one  cubic  foot  at 
sea  level 

(T       \  /  r 

T~  l)  =  58'8  \r» " 


HP  =  .00436  X  58.8  (~—  l\  ==  .2564  (~— 


the  same  result  we  obtained  in  a  former  example. 

INITIAL  TEMPERATURES. 

The  general  expression  for  the  work  of  compression 
being 


it  is  evident  that  so  long  as  -y-  remains  constant,  the 

power  to  compress  one  cubic  foot  of  the  same  gas  is  con- 
stant, but  inasmuch  as  the  temperature  of  the  mains  is 
practically  constant  and  about  60°  Fah.,  if  our  initial 
temperature  from  the  holder  should  happen  for  any 
reason  to  be  100°  Fah.,  as  it  was  entering  the  compres- 
sor, it  is  evident  that  the  compressor  must  make  an 
extra  number  of  revolutions  to  deliver  a  fixed  quantity 
into  the  mains  at  60°  Fah.  than  it  would  if  the  mains 
were  the  same  temperature  as  the  gas  in  the  holder, 
and  the  ratio  would  be  as  the  absolute  temperatures  or 

von  '  or  ^  Per  cen^  additional.  In  a  plant  where  250 
horsepower  is  used  in  compressing  the  gas,  this  would 
mean  a  saving  of  20  horsepower.  By  passing  the  gas 
through  a  cooler  before  it  reached  the  compressor 
would  correct  the  loss.  Inasmuch  as  little  water  is 
required  for  this,  and  the  water  is  in  no  wise  impaired 
for  other  purposes,  that  this  cooling  could  always  be 
done.  Vice  versa,  if  the  temperature  of  the  holder  was 


34          THE   COMPRESSION   AND   TRANSMISSION   OF   GAS 

lower  than  the  mains,  as  in  winter,  there  would  be  a 
corresponding  gain  and  some  of  the  otherwise  lost 
heat  of  compression  would  be  utilized  in  expanding 
the  gas  to  a  temperature  corresponding  to  the  main. 
In  the  long  run,  the  gain  might  balance  the  loss,  if  no 
cooling  were  done,  but  it  seems  a  business  proposition 
to  save  where  possible,  especially  where  it  costs  little 
or  nothing. 

TWO  STAGE  COMPRESSION. 

If  we  consider  the  general  equation  for  the  work  per- 
formed in  compressing  any  gas,  adiabatically 

L  =  JWC*(T  —  T°) 

we  note  that  the  only  variable  is  T,  the  final  tempera- 
ture, if  our  initial  temperature  remains  the  same.  In 
other  words,  the  difference  between  the  initial  and 
final  temperature  determines  always  the  power  ex- 
pended in  a  compressor,  just  as  it  does  the  power  given 
out  by  any  heat  engine.  It  is  evident,  then,  that  the 
lower  we  keep  the  final  temperature  the  less  power  it 
takes.  Water-jacketing  the  cylinders  accomplishes  but 
little,  probably  from  3  to  5  per  cent,  for  the  reason 
that  gases  being  such  poor  heat  conductors  that,  while 
they  are  rapidly  drawn  in  and  pushed  out  of  the  com- 
pressing cylinder,  there  is  not  time  for  the  heat 
to  radiate  through  the  cylinder  walls,  and  only  the 
portion  immediately  in  contact  with  the  cool  cylinder 
walls  suffers  any  reduction  of  temperature.  The  water 
jacket  keeps  the  cylinder  walls  cool  so  that  lubrication 
is  effective  and  is  valuable  for  that  reason  principally. 

Practically  speaking,  the  compression  is  adiabatic, 
or  even  greater  because  the  pressure  in  the  cylinder  is 
always  greater  than  the  receiver  on  account  of  the 
work  expended  in  forcing  the  gas  through  the  valve 
openings,  and  this  extra  heat  generated  overruns  the 
adiabatic  temperature  corresponding  to  the  receiver 
pressure. 

The  'water  jacket  being  ineffective,  the  device  of 
stage  compression  was  inaugurated,  where,  after  the 
gas  was  compressed  to  a  portion  of  the  final  pressure 


THE    COMPRESSION    AND    TRANSMISSION    OF    GAS          35 

in  a  cylinder,  it  was  discharged  into  an  intercooler,  its 
temperature  reduced  to  the  initial  and  then  compressed 
by  a  smaller  cylinder  to  the  final  pressure.  The  work 
was  found  to  be  a  minimum  when  the  final  temperature 
of  each  stage  was  the  same. 

If  we  represent  the  initial  pressure  by  P°  and  the 
final  by  P',  and  volumes  and  temperatures  similarly, 
we  shall  have,  using  our  general  formula  for  work  ex- 
pended, 

L  =  y^-P0  V°  ApS—  1 )  for  first  stage  and 

(T '  \ 

TJT —  1  I  f°r  second  stage. 
f          ) 

We  know  that  before  compression  P°  V°  must  equal 
P  F,  consequently  if  L  is.  desired  to  equal  L',  we  must 
have 


T'          /P'\  y-1  -i 

£_  =/ £_  I  — ,  or  reducing 


pj  =  p-  or  P2    =  P°  P'  or  P  =  l/P°  P' 

In  other  words,  to  make  the  work  in  two  stages 
equal,  and  to  have  the  work  a  minimum,  P,  the  inter- 
mediate pressure,  must  be  a  mean  proportional  be- 
tween the  initial  and  the  final  pressure,  the  volumes 
and  the  piston  areas  must  follow  the  same  law,  since 
we  naturally  make  the  strokes  alike. 

For  an  example,  let  us  take  80  pounds  final  gauge 
pressure : 


P  :   :   ]/P°  P'  or  v/14.7  x  94.7  =  37.31  absolute 
or  22.61  gauge  pressure.    This  makes 


36          THE    COMPRESSION    AND    TRANSMISSION    OF    GAS 


P_==37.31_2  54    and 

f-M-2'54 

and  inasmuch  as  these  pressure  ratios  are  the  same, 
the   work  expended  on   each   stage   will  be  the   same 
and  the  piston  ratio  will  be  2.54  also. 
We  found  for  the  standard  gas  that 

HP= .2564  (L  — 


Referring  to  Table  11,  we  find  when 
|i;  =2.54,  that  |j=  1.2624. 

Then  H  P  =  .2564  X  -2624  =  .06727  for  each  stage 
and  for  both  stages,  2  X  .06727  =  .13454  H  P. 

It  will  be  remembered  that  we  calculated  the  single 
stage  H  P  for  80  Ibs.  in  a  former  example  as  .1520. 
We  have  then  13.45  H  P  per  100  cu.  ft.  two  stage  against 
15.20  H  P  single  stage,  a  saving  of  13  per  cent  in  power. 

If  the  maintaining  of  a  low  temperature  is  any 
advantage  in  gas  compression,  we  have  a  temperature 
of  366°  Fah.  in  the  single  stage  compression  against 
195°  Fah.  in  the  two  stage,  a  remarkable  difference. 
Suppose  now  that  we  have  a  cylinder  having  an  area 
of  100  sq.  inches,  when  we  compress  to  80  Ibs.  the 
maximum  strain  is  8000  Ibs.,  if  the  compressor  is  single 
stage  and  4522  Ibs.  if  the  compressor  is  a  tandem  two 
stage,  a  remarkable  difference,  tending  to  show  that 
we  can  build  the  two  stage  compressor  very  much 
lighter  for  the  same  work. 

Another  point  in  favor  of  the  two  stage  compressor, 
it  has  a  greater  volumetric  efficiency.  A  piston  never 
delivers  from  a  cylinder  an  amount  of  gas  equal  to 
its  displacement,  because  clearance  spaces  are  filled 
with  gas  at  the  discharge  pressure,  which  expands  in 
the  return  stroke  of  the  piston  and  occupies  more  or 
less  space  according  to  the  ratio  of  compression  and 
the  amount  of  clearance.  The  greater  the  temperature 
of  compression,  the  hotter  the  piston  and  heads  and 


THE    COMPRESSION   AND   TRANSMISSION   OF   GAS          37 


^ 


38          THE    COMPRESSION    AND    TRANSMISSION    OF    GAS 

T 


S 
2 


III! 

JJJ  ! 

51  §  $i 


&fte 

\\ 


i» 


m 

$  va'N 


>»! 


Is 


«*"> 


g8 

m 


SK 


5»5> 


<o  « 


N  *.  *>j  1  h 


JK 


^^ 


"»™ 


NNN 


THE    COMPRESSION   AND    TRANSMISSION    OF    GAS          39 

valves  get,  and  the  less  weight  of  gas  enters  the  cyl- 
inder on  account  of  the  clearance  expansion.  There  are 
other  losses  which  need  not  be  mentioned  here,  but  these 
two  are  sufficient  to  make  the  volumetric  efficiency  of 
single  stage  compressors  at  80  Ibs.  average  about  75 
per  cent. 

It  will  be  readily  seen  that  the  initial  cylinder  of  a 
two-stage  machine  at  80  Ibs.  will  have  its  clearance 
losses  divided  by  2.54,  because  that  will  be  the  relative 
ratio  of  pressures  and  the  temperature  losses  in  pro- 
portion to 

because  that  is  the  absolute  temperature  ratio. 

These  combined  will  make  the  average  two-stage  com- 
pressor good  for  90  per  cent  volumetric  efficiency— 
in  other  words,  15  per  centr  better  than  a  single  stage. 
One  can,  therefore,  afford  to  pay  at  least  15  per  cent 
more  for  a  two-stage  machine  than  for  a  single-stage 
machine,  the  intake  cylinders  being  the  same  size,  and 
this  extra  15  per  cent  will  nearly,  or  sometimes  quite, 
pay  for  the  difference  in  price. 

It  is  evident  from  the  calculations  we  have  made 
that  the  efficiency  of  a  two-stage  machine  over  the 
single  stage  increases  directly  as  the  pressure  ratios 
increase,  and  inasmuch  as  altitude  increases  pressure 
ratios,  it  is  evident  that  the  higher  the  altitude  the  more 
urgent  becomes  the  necessity  for  using  the  two-stage 
machines,  and  at  altitudes  above  3000  feet  it  is  prac- 
tically imperative. 

Theoretically,  an  infinite  number  of  stages  world 
give  isothermal  compression,  but  practically  the  losses 
involved  in  driving  the  gas  through  too  many  cylin- 
ders and  valves  would  offset  this  gain,  and  we  can  con- 
sider that  two  stages  will  probably  be  the  limit  for  all 
ordinary  purposes. 

ALTITUDE  COMPRESSION. 

We  found  that  it  took  the  same  power  to  compress 
one  cubic  foot  of  gas  at  any  temperature  to  the  same 
final  pressure,  provided  the  initial  pressures  were  the 


40 


THE    COMPRESSION   AND    TRANSMISSION    OF    GAS 


same,  and  it  naturally  followed  that  it  took  more 
power  to  compress  the  same  weight  at  higher  tempera- 
tures, because  there  would  be  a  larger  volume  and  the 
piston  would  have  to  make  more  strokes. 

Altitude  acts  like  an  increase  of  temperature  in  les- 
sening the  density  of  a  gas,  but  it  introduces  another 
element,  viz.,  change  of  initial  pressure,  so  that  as  we 
reach  higher  altitudes  the  pressure  ratio  is  constantly 
increasing,  which  means,  of  course,  that  the  tempera- 
ture of  compression  is  increasing  and  more  work  per 
unit  of  gas  weight  is  being  done,  but  the  weight  is 
constantly  decreasing  as  we  ascend,  and  the  combina- 
tion of  these  results  is  that  while  it  takes  less  work  to 
discharge  any  given  cylinder  full  of  gas  at  an  altitude, 
the  increased  number  of  strokes  necessary  to  com- 
press a  weight  equivalent  to  a  given  sea  level  volume  is 
considerably  greater. 


Table  17. 


July,  1905 


**  e    : 

2,  6        ; 

o    : 

^  S        ^ 

jj 

Altitude. 

P 
po 

Y 

Y—  1 

po 

F° 

£o-l 

Icl 

ii!j 

S 

£ 

B 

II! 

|l  ii 

'S 
'S 

3 

o 

Sea  level.. 

6.44 

4 

14.7 

1XH4 

.593 

5020 

5020 

14.7 

80 

10,000  ft.. 

9.47 

4 

10. 

1X144 

.753 

4337 

6375 

10. 

80 

T  =  390°  Fah.  at  sea  level. 

T  =  450°  Fah.  at  10,000  feet  altitude. 

Table  17,  shows  a  comparison  between  compressing 
gas  at  a  sea  level  and  at  10,000  feet  altitude  single  stage 
compression.  The  columns  3  to  6,  inclusive,  comprise 
the  components  of  the  general  formula  for  compressing 
gas,  and  it  is  interesting  to  note  the  variable  quantities. 
It  will  be  seen  that  while  one  cubic  foot  of  the  altitude 
gas  requires  less  power,  the  increased  volume  necessary 
to  produce  a  common  result  makes  it  require  25  per 
cent  more  power. 

It  will  also  be  noted  that  the  final  temperature  is 


THE    COMPRESSION    AND    TRANSMISSION    OF    GAS          41 

quite   high   in   comparison   to   sea   level   compression, 
which  speaks  loudly  for  two-stage  compression. 

FLOW  OF  GAS  IN  PIPES. 

After  reading  the  report  of  the  committee  on  "The 
Flow  of  Gas  in  Pipes,  '  '  for  the  Ohio  Gas  Light  Associa^ 
tion,  as  published  in  the  American  Gas  Light  Journal, 
April  24,  1905,  the  general  impression  would  be  that 
the  formulas  were  not  sufficiently  reliable  to  be  of  great 
service,  because  there  was  a  variation  in  the  results  of 
a  given  problem  of  from  1  to  200  per  cent.  It  would 
seem,  however,  that  six  formulas  out  of  the  nine  do 
not  vary  15  per  cent,  and  the  three  most  frequently 
used  do  not  vary  2%  per  cent. 

If  we  should  accept  the  largest  of  these  three,  called 
the  Pittsburg  formula,  we  wrould  probably  not  be  far 
wrong,  and  particularly  as  the  results  do  not  differ 
greatly  from  those  obtained  by  using  Cox's  computer, 
and  I  am  informed  by-  those  who  have  used  the  com- 
puter that  it  is  perfectly  safe. 

Again,  the  variation  in  the  areas  of  those  pipe  sizes 
most  likely  to  be  used  are  much  more  considerable  than 
the  variations  of  any  of  the  six  formulas  above  referred 
to.  Thus,  taking  the  commercial  sizes  of  pipe  from 
1"  to  6",  the  average  variation  between  the  areas  of 
each  size  is  35  per  cent. 

If  we  therefore  make  a  practice  of  using  the  pipe 
that  is  the  nearest  size  larger  than  our  calculations, 
we  shall  have  an  ample  safety  factor. 

For  air  we  have  been  using  a  formula  developed  by 
Mr.  J.  E.  Johnson,  Jr.,  and  published  in  the  American 
Machinist  July  27,  1899. 

-0006     *  L 


P'  =  absolute  initial  pressure. 
P"  =  absolute  final  pressure. 
Q  =  free  air  equivalent  in  cubic  feet  per  minute. 
L  =  length  of  pipe  in  feet. 
d  =  diameter  on  pipe  in  inches. 
Practical  results  from  this  formula  show  that  it  is  a 


42          THE   COMPRESSION   AND   TRANSMISSION   OF   GAS 


little   too   liberal,    and  that   P'2  —  P"2  =  > 

Us 

would  be  nearer  the  results. 

The  Pittsburg  gas  formula  reduces  to  the  same  value 
when  the  proper  substitutes  are  made  for  the  relative 
specific  gravities  of  gas  and  air. 

Inasmuch  as  the  specific  gravity  of  gas  is  always  re- 
ferred to  air  as  1,  it  seems  right  that  our  gas  formula 
should  refer  to  air  and  a  co-efficient  used  for  each  gas. 

The  velocity  of  different  gases  through  a  pipe  varies 
inversely  as  the  square  root  of  their  densities,  or  what 
amounts  to  the  same  thing,  their  specific  gravities  or 
weights  compared  to  air,  then  the  velocities  will  vary 
as 


±-    or   VG 

Where  G  is  the  specific  gravity  of  gas. 
Prefixing  this  to  our  original  equation,  we  have  in 
general,  for  any  gas, 

p/2  _  p//2  _  .o005i/£  x  <~^ 
Or 


44.72       |P'2  —  P"2  X 


4^-il         I 

I/  a  \ 


L 

Inasmuch  as  certainly  for  some  considerable  time 
crude  oil  gas  will  be  most  extensively  used  by  members 
of  this  Association,  let  us  substitute  in  the  above  form- 
ula the  value  of  the  largest  probable  specific  gravity, 
viz.,  .49,  and  we  have  1/749  ~  -1  and 

p/2  __  p//  _  00035  9L^.  (A) 

d5 
Or 

|  D/2  P//2     V/     fJo 

Q  =  53.45  J"        JL- 

Q  is  in  cubic  feet  per  minute  rather  than  per  hour, 
because  all  compressors  are  so  rated. 

Table  12  gives  values  of  P'2  -  -  P"2  for  100  feet  of 
various  sized  pipes  and  quantities  will  be  found  conve- 
nient for  figuring  gas  flows  in  pipes.  The  values  are 
calculated  from  equation  (A). 


THE    COMPRESSION   AND    TRANSMISSION    OF    GAS          43 


JUL.Y  /3O5 


C/fL  . 


Jo 


A  Ke3 


£/l  ff/x. 


44          THE    COMPRESSION    AND    TRANSMISSION    OF    GAS 

Example  I — 

1000  cubic  feet  per  minute  of  gas  at  90  pounds  gauge 
pressure  is  discharging  into  a  4"  pipe  26,000  feet  long. 
Required  the  terminal  pressure. 
P'2  =  10962.     (Table  13) 
P's  __  p"2  =  35.04  for  100  feet.     (Table  12) 
Multiplying  by  260  for  26,000  feet 
P'2  __  p"2  —  9HO. 

p//2  =  p/2  _   (p/2_p//2)  =  i0962  _  9110  =  1852. 
P"2  =  1852.    P"  =  28  pounds. 

Example  II— 

A  pipe  line  3"  diameter  and  11,000  feet  long.  Re- 
quired to  find  the  quantity  of  gas  that  will  be  delivered 
at  a  terminal  pressure  of  1  pound,  the  initial  pressure 
being  40  pounds. 

P'2  =  2992.     (Table  13) 

P"2  =  279     (Table  13) 

p/2  _  p//2  =  2713  for  11,000  feet  of  pipe  or  24.6 
for  100  feet. 

Referring  to  Table  12,  we  find  value  of  23.70  for  420 
cubic  feet  per  minute. 

Example  III— 

A  pipe  line  is  11,000  feet  long  and  4"  diameter.  The 
equivalent  of  1000  cubic  feet  is  wanted  at  the  end  of 
the  line  at  10  pounds  pressure.  What  must  be  the 
initial  pressure? 

p/2  _  p//2  _  35  04  for  100  feet  (Table  12).  Multiply- 
ing by  110  we  have 

p/2  _  p//2  _  3854  for  ii?000  feet. 

P"*  =  610.     (Table  13) 

p/2  =  p//2  +    (p/2  _  p//2)  _  3854  +  610  ==  4464 

P'  ==  j/4464. 

Referring  to  Table  13  we  find  52  pounds  gauge  pres- 
sure to  be  the  initial  pressure. 

Example  IV— 

The  equivalent  of  200  cubic  feet  per  minute  is  to  be 
put  through  a  pipe  53,000  feet  long.  The  initial  pres- 


THE    COMPRESSION    AND    TRANSMISSION    OF    GAS          45 

sure  is  20  pounds.    The  final  pressure  must  be  6  pounds. 
What  will  be  the  size  of  the  pipe  ? 
P'2  =  1204.     (Table  13) 

P"2  =  428.     (Table  13) 

p/2  _  P"2  =  776  for  53?000  feet  of  pipe  or  l  464 

per  100  feet.    Referring  to  Table  12,  we  find  4"  to  be 
the  proper  size. 

SOME  CORROBORATIONS. 

Table  15  gives  at  Figure  1  a  card  from  the  gas  cylin- 
der of  a  compressor  at  Fresno,  compressing  crude  oil 
gas  at  a  pressure  of  27  pounds  gauge. 

If  we  draw  the  line  of  27  pounds'  pressure  and  take 
the  M  E  P  with  a  planimeter,  following  the  curve  A  B 
and  the  straight  lines  B  C  —  CD  and  A  B,  we  shall 
have  the  M  E  P  of  a  perfect  card  following  the  actual 
compression  line.  This  M  E  P  we  find  to  be  17.4 
pounds,  using  the  y  which  we  found  for  Fresno  gas, 
the  adiabatic  HEP  for  27  pounds  =  17.58,  making  a 
good  check  on  our  values. 

Figures  2  and  3  are  from  a  compressor  pumping 
natural  gas  at  Anderson,  Indiana,  each  having  an  in- 
take pressure  of  11  pounds — drawing  lines  of  50 
pounds'  pressure  at  Figure  2  and  60  pounds  at  Figure 
3,  and  taking  the  M  E  P  in  the  same  way  that  we  did 
in  Figure  1,  we  find  that  the  M  E  P  for  Figure  2  is  26 
pounds,  and  for  Figure  3,  30  pounds. 

Using  the  value  of  y  which  we  developed  for  natural 
gas  and  calculating  the  adiabatic  M  E  P,  we  find  they 
are  26.30  and  30.85  pounds,  respectively,  a  very  satis- 
factory check,  and  from  these  we  may  fairly  conclude 
that  our  theories  and  formulas  are  reasonable. 

It  will  be  noted  that  the  line  of  the  compressor  curve 
is  very  near  the  adiabatic,  even  though  the  compressors 
were  making  but  60  to  70  revolutions  per  minute.  An 
air  card  would  show  at  least  double  the  separating 
space. 

This  would  appear  to  show  that  the  jackets  were 
doing  but  very  little  good,  and  possibly  because  illu- 
minating gas  may  be  a  much  poorer  conductor  of  heat 
than  air. 


46 


THE    COMPRESSION    AND    TRANSMISSION    OF    GAS 


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THE    COMPRESSION   AND    TRANSMISSION   OF   GAS          47 

The  line  of  compression  comes  so  near  the  adiabatic 
that  we  may  well  call  the  compression  adiabatic^  for 
safety  in  our  calculations — but  while  the  M  E  P  adia- 
batic for  any  pressure  represents  the  greatest  possible 
power  required  to  compress  a  gas,  a  still  greater  power 
must  be  applied — for  example  look  at  the  Fresno  card, 
Figure  1,  Table  14 — the  area  above  the  27-pounds  line 
represents  work  done  in  overcoming  the  inertia  of  the 
outlet  valves  in  pushing  the  gas  into  the  main,  and  this 
area  will  be  greater  or  less  depending  upon  the  valve 
area  and  the  size  of  the  discharge  openings  and  the 
piston  speed.  It  will  also  be  noted  that  there  is  an  area 
representing  suction  work  below  the  line  A  D,  not- 
withstanding that  the  gas  has  a  4"  water  pressure  at 
holder.  This  probably  indicates  that  the  pipes  from 
the  holder  to  the  compressor  are  too  small. 

Now,  if  we  run  a  planimeter  over  the  actual  area  of 
the  card,  we  find  that  the  real  M  E  P  is  19.4,  or  about 
10  per  cent  greater  than  the  adiabatic,  and  this  agrees 
quite  well  with  ordinary  air  practice,  where  a  safe  rule 
for  single-stage  work  is  to  take  the  M  E  P  at  10  per 
cent  above  the  adiabatic  and  the  two-stage  M  E  P  the 
same  as  the  adiabatic.  Slow  speed,  well-constructed 
compressors  will  do  somewhat  better,  but  it  is  well  to 
calculate  on  the  average  type. 

Now,  for  brake  power  to  be  delivered  to  a  gas  com- 
pressor, we  have  to  allow  a  mechanical  efficiency  of  the 
compressor  at  not  to  exceed  85  per  cent,  so  that  this  15 
per  cent  loss  combined  with  the  10  per  cent  loss  in  the 
cylinder  points  to  the  fact  that  we  should  add  26^/2 
per  cent  to  the  adiabatic  H  P  for  the  brake  power  re- 
quired. 

The  steam-engine  cards  on  the  Fresno  compressor 
show  an  M  E  P  reduced  to  the  size  of  the  air  cylinder 
of  20.75  pounds,  or  20  per  cent  higher  ,than  the  adia- 
batic air  M  E  P,  but  this  compressor  had  a  Meyer  cut- 
off, which  helped  its  economy  considerably. 

Referring  to  Table  9,  column  17,  gives  the  formula 
for  computing  the  power  to  compress  one  cubic  foot 
of  the  gas  at  sea  level  and  60°  Fah.  If  the  calculation 


48          THE    COMPRESSION   AND    TRANSMISSION    OF    GAS 

be  made  it  will  be  noted  that  it  takes  practically  the 
same  power  to  compress  one  cubic  foot  of  any  of  these 
gases,  consequently  Table  19  may  be  used  generally. 

In  conclusion  your  attention  is  called  to  Table  19, 
which  contains  in  convenient  form  the  results  which  we 
have  obtained,  and  which  it  is  hoped  you  will  find  very 
helpful  in  considering  thermodynamic  questions  re- 
garding the  standard  illuminating  gas  made  from 
crude  oil. 


SOME    ECONOMICS    IN    HIGH    PRESSURE    GAS 
TRANSMISSION. 

By  EDWARD  A.  MX. 

Mr.   President    and    Fellow    Members    of    the   Pacific 
Coast  Gas  Association. 

Last  year  I  had  the  pleasure  of  presenting  for  your 
consideration  a  paper  entitled  "The  Compression  and 
Transmission  of  Illuminating  Gas,"  in  which  the  gen- 
eral theory  of  the  subject  was  discussed,  and  methods 
shown  for  calculating  the  specific  heats  of  various  gas 
mixtures,  the  heat  developed  by  compression,  and  the 
power  required.  Also  various  losses  in  pressure  and 
power  in  pushing  gas  through  pipes,  and  some  co- 
efficients for  all  this  data,  so  that  one  could  approxi- 
mate, with  some  degree  of  certainty,  the  various  ele- 
ments of  a  practical  plant.  The  length  of  the  paper 
did  not  permit  of  bringing  it  to  such  a  closing  that 
those  not  caring  for  the  theoretical  part  could  readily 
solve  some  everyday  problems  pertaining  to  a  high- 
pressure  gas  transmission.  It  is,  therefore,  the  intent 
of  this  paper  to  briefly  supply  this  deficiency,  and  at 
the  same  time  to  introduce  the  element  of  cost  of 
compressing  gas  and  a  method  for  determining  the 
proper  size  of  pipes  so  that  the  gas  engineer  may  be 
able  to  readily  and  easily  arrive  at  the  essential  ele- 
ments in  his  problem. 

Curve  sheet  No.  20  has  been  constructed  for  this 
purpose  and  conditions  as  general  as  possible  have 
been  assumed,  and  in  order  that  it  may  be  easy  for 
anyone  to  construct  a  similar  curve  sheet  for  other 
conditions,  the  method  of  making  it  may  well  be 
explained.  Inasmuch  as  gas  is  generally  sold  and 
handled  by  the  1,000  cubic  feet,  it  seems  proper  to 
make  that  the  basis  for  quantity,  and  one  cent  per 
kilowatt  hour  seems  a  natural  base  from  which  to  cal- 
culate the  cost  of  power,  and  should  anyone  have  steam 
power  or  power  other  than  electric,  it  is  a  simple  mat- 
ter to  convert  it  to  kilowatt  hours. 


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rf«*4A«*««4 


HIGH    PRESSURE    GAS    TRANSMISSION  51 

If  a  kilowatt  hour  costs  one  cent,  a  horsepower  will 
cost  practically  three-quarters  of  a  cent,  one  horse- 
power hour=60x33,000  or  1,980,000  foot  pounds  for 
three  quarters  of  a  cent,  one  cent  would  therefore  pro- 


000 

duce  ::p-  =2,640,000  foot  pounds. 
.  (  o 

Eight  horsepower  equals  264,000  foot  pounds,  con- 
sequently every  8  horsepower  will  cost  1-10  of  a  cent 
per  minute.  This  gives  us  the  basis  for  our  curve,  for 
if  we  lay  out  our  sheet  in  equal  divisions  of  any  size 
and  call  each  one  along  the  vertical  line  8  horsepower, 
we  can  also  make  each  division  represent  1-10  of  a 
cent,  and  each  horizontal  division  we  can  conveniently 
call  10  pounds. 

If  you  will  now  refer  to  Table  19,  the  last  table 
which  I  read  at  our  last  meeting,  it  will  be  possible 
to  construct  the  curve,  remembering  that  the  table 
is  constructed  for  100  cubic  feet  per  minute,  the  horse- 
power therein  contained  must  be  multiplied  by  10,  for 
the  1000  feet  capacity  we  are  now  considering. 

Take,  for  example,  50  pounds  gauge  pressure,  the 
brake  horse-power  required  for  100  cubic  feet  is  14.56, 
and  for  1000  would  be  145.6.  Where  the  vertical  line 
indicating  50  pounds  meets  the  horizontal  line  drawn 
from  145  horse-power  will  be  a  point  on  the  curve. 
Similarly  other  points  can  be  made,  and  joining  the 
points  together,  we  shall  have  a  cost  and  power  curve 
combined  which  will  be  very  useful  in  our  calculation. 

I  have  constructed  two  of  these  curves,  A  and  B. 
A  is  the  curve  of  single  stage  compression  and  B  for 
two-stage  compression.  Single  stage  is  rarely  used 
beyond  100  pounds  pressure,  nor  two  stage  below  90 
pounds  pressure.  You  will  note  quite  a  difference  in 
favor  of  two-stage  compression.  For  example,  at  100 
pounds  pressure  it  costs  2.35  cents  per  1000  for  two- 
stage  and  2.75  cents  for  single-stage.  In  even  a  small 
plant  using  50,000,000  feet  per  year,  the  difference 
would  be  $200  per  annum,  which  is  well  worth  saving. 

The  two-stage  curve  may  be  readily  •  constructed 
from  the  single-stage  curve  by  remembering  that  the 


52  SOME    ECONOMICS    IN 

intermediate  absolute  pressure  between  the  stages  is 
a  mean  proportional  between  the  initial  and  final  abso- 
lute pressure,  and,  inasmuch  as  it  takes  the  same 
power  for  each  stage,  if  we  double  the  power  required 
for  the  first  stage  we  shall  have  the  desired  results, 
thus  the  intermediate  gauge  pressure  for  200  pounds 
pressure  will  be  41  pounds.  We  note  from  the  single- 
stage  curve  that  41  pounds  requires  128  horse-power, 
consequently  twice  this  is  256  horse-power,  which,  laid 
out  on  our  curve  sheet  on  the  200-pound  vertical  line, 
will  give  us  the  point  N  on  the  two-stage  curve,  and  so 
on  for  other  points  to  complete  the  curve. 

It  must  be  understood  that  the  horse-powers  are  for 
1000  cubic  feet  per  minute,  and  the  cost  will  be  per 
1000,  and  if  you  wish  to  eliminate  the  element  of  time 
just  multiply  the  horse-power  by  33,000  and  the  re- 
sult will  be  the  foot  pounds  to  compress  1000  cubic 
feet  of  gas,  and  independent  of  time. 

If  power  costs  more  or  less  than  1  cent  per  kilowatt 
hour,  or  the  quantity  to  be  compressed  is  greater  or 
less  than  1000  cubic  feet  per  minute,  the  results  may 
be  read  from  the  curve  by  simply  using  a  correspond- 
ing proportion,  for  example : 

The  curve  shows  that  1000  cubic  feet  can  be  com- 
pressed to  20  pounds  gauge  pressure  at  the  cost  of  1 
cent,  it  follows,  therefore,  that  2000  cubic  feet  can  be 
compressed  to  20  pounds  for  2  cents,  or  if  power  costs 
2  cents  per  kilowatt  hour  instead  of  1  cent,  then  only 
one-half  the  quantity  can  be  compressed  for  1  cent,  or 
double  the  quantity  if  power  costs  but  y2  a  cent  a 
kilowatt  hour.  This  method  of  proportion,  however, 
does  not  apply  to  the  matter  of  pressure,  for  you  will 
note  that  while  a  cost  of  1  cent  gives  20  pounds  pres- 
sure, a  cost  of  2  cents  gives  58  pounds  pressure,  and  a 
cost  of  l/2  a  cent  gives  only  8  pounds  pressure.  In 
other  words,  it  costs  just  as  much  to  compress  gas 
from  0  to  8  pounds  as  it  does  from  8  to  20  pounds, 
and  just  as  much  to  compress  from  0  to  20  pounds  as 
from  20  to  58  pounds.  It  would  be  well  right  here 


HIGH    PRESSURE    GAS   TRANSMISSION  53 

to  consider  this  fact,  for  it  has  a  great  bearing  on  high- 
pressure  transmission. 

If  it  was  found,  for  example,  that  it  was  costing  1 
cent  per  1000  to  deliver  gas  through  a  certain  pipe 
at  20  pounds  pressure,  and  it  became  necessary  to 
double  the  pressure  in  order  to  supply  an  increased  de- 
mand, the  gas  company  might  consider  it  inadvisable 
because  it  might  double  the  cost.  Consulting  the  curve, 
it  will  be  seen  that  the  cost  for  compressing  at  40 
pounds  pressure  is  only  1.6  cents  per  1000  cubic  feet 
instead  of  2  cents,  as  may  be  imagined,  and  this  fact 
might  justify  the  increased  pressure,  and  the  higher 
the  pressure  the  more  the  seeming  disproportion. 

From  the  curve  take  a  geometrical  progression  of 
gauge  pressure,  5-10-20-40-80-160-320,  and  we  note  the 
corresponding  costs  of  compression  for  1000  cubic  feet 
to  be,  in  cents,  .3-.575-1.00-1.6-2.4-3.-3.9,  in  other  words, 
while  the  pressure  from  5  to  320  has  increased  sixty- 
four  times,  the  cost  of  compression  has  increased  but 
thirteen  times. 

It  must  not,  however,  be  hastily  inferred  that  be- 
cause of  this  decreasing  power  ratio  that  it  is  econom- 
ical to  compress  at  high  pressure,  because  it  may  not 
be  so  and  depends  upon  the  amount  of  gas  to  be 
pumped,  for  while  the  rate  for  1000  may  be  small  and 
make  no  material  difference  where  a  small  quantity  is 
pumped,  with  a  large  quantity  the  total  amount  of 
the  yearly  cost  of  pumping  may  exceed  so  materially 
the  interest  and  depreciation  on  a  larger  pipe  using  a 
lower  pressure  that  the  latter  installation  will  be 
deemed  preferable. 

The  question  of  whether  a  large  pipe  and  small 
pressure,  or  a  small  pipe  and  high  pressure  shall  be 
used  is  simply  a  matter  of  equating  the  relative  costs 
of  pumping,  together  with  the  interest  and  deprecia- 
tion on  the  plant,  and,  with  the  curve  given  herewith, 
it  may  be  easily  determined,  and  it  is  to  show  how  to 
make  this  determination  quite  accurately  and  simple 
that  I  have  written  this  paper.  It  has  seemed  to  me 
that  the  cost  of  attendance,  buildings,  laying  out 


54  SOME   ECONOMICS   IN 

pipe,  etc.,  for  any  one  problem  may  be  neglected,  for 
while  it  is  an  item  of  cost,  it  will  be  practically  the 
same  for  whatever  pipe  you  may  select.  We  can  then 
make  our  comparison,  using  the  market  cost  of  pipe 
and  the  cost  of  power  only,  to  which  may  be  added  the 
other  costs  after  the  size  of  the  pipe  is  determined, 
in  order  to  give  the  total  cost  per  1000  for  handling 
the  gas. 

The  power  and  cost  curves,  as  constructed,  can  be 
called  " Standard"  and  white  prints  made  from  it, 
and  upon  these  white  prints  the  pipe  curves  laid  out, 
as  will  be  shown,  and  this  same  white  print  can  be 
used  in  all  cases.  Let  us  then  take  two  examples,  one 
for  small  quantity  and  one  for  large  quantity,  and 
before  starting  at  it  let  us  make  a  general  standard 
formula,  which  will  simplify  many  of  the  calculations. 

No  plant  will  pump  less  than  10,000,000  cubic  feet 
per  year,  which  is  about  1200  per  hour,  or  at  a  less 
distance  than  10,000  feet,  consequently  take  for  a 
basis : 

10,000,000  feet  per  year=a 
10,000  feet  of  pipe=Z> 
1  cent  per  foot  cost  of  pipe 
10%   per  annum  interest  and  depreciation 
on  the  pipe. 

Then  equating  these  quantities  we  will  find  that  the 
pipe  cost  C  for  1000  cubic  feet  of  gas  will  be  1/10  of  a 
cent. 

For  any  other  quantity  Q,  and  length  of  pipe  L,  and 
price  of  pipe  P,  we  shall  have: 

Pipe  cost  per  1000  C  =  ~?  X  ^ 
o\2        1U 

Example  1—50,000,000  cubic  feet  per  year,  or  6000 
per  hour,  50,000  feet  of  pipe,  and  power  to  cost  i/o  cent 
per  kilowatt  hour,  substituting  in  our  formula 

La       P  50,000  X  10,000,000      JP  =_JP 

bQ      10 W  10,000  X  50,000,000      10"    10 

That  is  to  say  that  whatever  size  pipe  we  select  the 

pipe  cost  per  1000  cubic  feet  of  gas  will  be  1/10  the 

market  cost  of  pipe  per  foot. 


HIGH    PRESSURE    GAS   TRANSMISSION  55 

Taking  the  market  prices  of  to-day,  then  if  we  should 
use: 
1%  pi?6?  cost  per  1000  cubic  feet,  equals  .........  559 

iy2  pipe,  cost  per  1000  cubic  feet,  equals  .........  67 

2  pipe,  cost  per  1000  cubic  feet,  equals  ......  .  .    .894 

21/k  pipe,  cost  per  1000  cubic  feet,  equals.  .......  1.429 

3  pipe,  cost  per  1000  cubic  feet,  equals  ........  1.875 

Having  thus  blocked  out  the  matter  of  pipe,  we  must 

find  what  pressure  it  is  necessary  to  use  to  pump  the 
gas  through  these  various  sized  pipes,  assuming  al- 
ways that  the  terminal  pressure  shall  be  one  pound 
gauge. 

You  will  remember  that  we  developed  a  formula  in 
my  paper,  read  last  year,  which  may  be  used  here  with 
accurac. 


is  the  difference  between  the  squares  of  the  initial  and 
final  absolute  pressures. 

Q  is  the  quantity  of  gas  in  cubic  feet  per  minute. 

L  is  the  length  of  pipe,  in  feet. 

D  is  the  pipe  diameter,  in  inches. 

Substituing  in  this  equation  the  elements  in  our  prob- 
lem, we  have  : 

_35X100X  100X50,000  _  175,000 

Fl~~  ~100,00<fXd~  Fl~  d«- 

Then  P/2  XP22  equals  for 
1%  pipe    .......................  57,370 

1V2  pipe   .......................  23,000 

2  pipe   ........................   5,500 

2%  pipe   .......................   1,800 

3  pipe    .................  .  .....      700 

P2being  our  final  pressure  lib.  or  15.7  Ibs.  absolute  makes 
P./2=246,  and  remembering  that  Pl2—?^=(Pl2—?l2Jr 

P/2 
we  have 

\l/4  pine  P./2  absolute=57,616  then  Pl  gauge=225  Ibs. 

11/2    "  "  "      =23,246     "  "  "  -140  " 

2  "  "  "      =  5,746     "  "  "  ==  61  " 
2l/2   »  "  "      =  2,046     "  "  "  =  31  " 

3  "  "  946     "  M  "  --  16  " 
Now  we  are  ready  to  put  all  this  on  our  curve  sheet 

in  order  that  we  may  have  a  graphic  representation  of 


0  SOME    ECONOMICS   IN 

the  situation.  The  power  cost  curve  is  on  a  basis  of  1 
cent  per  kilowatt  hour;  if,  therefore,  we  plot  any 
other  costs  on  this  standard  sheet,  they  must  be  in- 
creased or  diminished  by  the  ratio  the  actual  power  cost 
bears  to  the  standard  power  cost  of  1  cent  per  kilo- 
watt. Our  problem  calls  for  a  power  cost  of  %  cent 
per  kilowatt  hour,  consequently  we  must  plot  in  our 
pipe  costs  at  double  their  real  amount,  for  the  standard 
power  cost  curve  is  double  the  cost  stated  in  our  prob- 
lem. Take,  then,  the  2-inch  pipe.  We  have  found  the 
pipe  cost  to  be  .894,  which,  multiplied  by  2,  equals 
1.798,  and  the  initial  pressure  required  is  61  pounds. 
If  then  we  lay  off  the  point  P  on  the  61  pounds  pres- 
sure ordinate  and  opposite  the  cost  Line  1.798  or  1.8,  we 
shall  have  the  two-inch  pipe  established,  and  similarly 
we  can  establish  the  other  points,  and  joining  them  by 
lines  we  have  a  pipe  cost  curve  S.  C.,  which  shows 
the  cost  of  pipe  per  1000  cubic  feet  of  gas  for  all  sizes 
from  1%  to  3  inch.  It  is  evident  that  the  total  cost 
of  pumping  the  gas  is  the  pipe  cost  plus  the  power  cost, 
consequently  if  we  add  the  pipe  cost  curve  to  the  power 
cost  curve,  we  shall  have  a  curve  of  total  cost.  Take 
the  2-inch  pipe  once  more,  and  with  dividers  measure 
off  the  distance  W.  P.,  the  cost  of  the  pipe  line,  and 
add  it  to  the  line  W.  Y.,  which  is  the  power  cost,  and  we 
have  the  point  T.,  as  the  result  of  the  addition.  Do  the 
same  for  the  other  sizes  of  pipe,  join  these  points  by 
lines,  and  we  have  the  curve  D.  F.  as  the  final  result, 
showing  the  combined  power  and  pipe  cost  per  1000 
cubic  feet  of  gas  for  all  the  sizes  of  pipe  under  consid- 
eration. 

It  is  evident  at  a  glance  that  the  2-inch  pipe  shows 
the  least  cost,  consequently  the  solution  of  our  prob- 
lem is  2-inch  pipe  at  60-pound  initial  pressure.  Total 
cost  per  1000  3.6  cents  on  a  basis  of  1  cent  per  kilowatt 
hour,  but  as  our  power  costs  %  a  cent  per  kilowatt 
hour,  the  total  cost  is  1.8  cents  per  1000  cubic  feet  of 
gas. 

The  graphical  method  is  very  satisfactory,  for  one 
can  see  at  a  glance  the  relations  between  the  various 


HIGH    PRESSURE    GAS    TRANSMISSION  57 

elements,  for  example :  You  will  note  that  it  cost  prac- 
tically the  same  to  pump  this  gas  through  a  l^-inch 
pipe  at  225  pounds  pressure  or  a  li/o-inch  pipe  at  140 
pounds  pressure  or  the  3 -inch  pipe  at  16  pounds  pres- 
sure, which  is  interesting. 

PROBLEM    TWO. 

Take  5000  cubic  feet  per  minute,  300,000  per  hour  or 
2,500,000,000  per  year,  through  the  same  length  pipe 
as  in  our  former  problem,  and  at  the  same  power  cost 
per  kilowatt  hour.  The  pipe  being  the  same  length 
and  the  quantity  fifty  times  greater.  We  will  have 

P  P 
for  pipe  cost  per  1000  where  we  had       formerly, 

and  taking  pipe  casing  prices  up  to  12  inches,  which 
was  the  largest  I  could  get,  and  assuming  them  above 
that  size  simply  for  illustration,  we  find  the  following 
prices  per  1000  feet  of  gas  for  these  pipes : 

10  inch  equals    .25  cent  per  1000  cubic  feet  of  gas. 

12  inch  equals    .30  cent  per  1000  cubic  feet  of  gas. 

14  inch  equals    .40  cent  per  1000  cubic  feet  of  gas. 

16  inch  equals    .60  cent  per  1000  cubic  feet  of  gas. 

18  inch  equals    .80  cent  per  1000  cubic  feet  of  gas. 

20  inch  equals  1.00  cent  per  1000  cubic  feet  of  gas. 
and  the  respective  pressures  necessary  to  force  the  gas 
through  these  pipes  to  be : 

10  inch  equals  53  pounds.  16  inch  equals  12  pounds. 

12  inch  equals  26  pounds.  18  inch  equals    8  pounds. 

14  inch  equals  20  pounds.  20  inch  equals    4  pounds. 

Multiply  the  pipe  cost  by  2  and  transferring  those 
quantities  to  our  curve  sheet,  precisely  as  we  did  in  the 
other  problem,  we  have  a  pipe  curve,  E,  showing  the 
pipe  cost  per  1000  cubic  feet,  and  adding  this  to  the 
power  curve,  we  have  the  final  result  in  the  curve  F, 
which  shows  this  rather  surprising  fact,  that  the  12- 
inch  pipe  is  the  best,  and  that  it  will  carry  this  gas  at 
26  pounds  pressure,  and  at  a  cost  of  8/10  of  a  cent  per 
1000  for  power  and  pipe. 

The  curve  shows  also  that  the  gas  can  be  put  through 
the  10-inch  pipe  at  53  pounds  pressure  at  the  same 


58  SOME    ECONOMICS   IN 

cost  as  through  the  20-inch  pipe  at  4  pounds  pressure. 
I  believe  this  graphical  method  will  be  of  service  to 
you,  particularly  as  these  curve  sheets  can  be  filed 
away  and  always  used  for  quick  reference,  for  the  eye 
can  take  in  the  whole  relative  situation  at  a  glance.  I 
believe  this  method  of  handling  the  subject  removes  it 
from  the  realm  of  speculation,  and  makes  an  orderly 
comparison  of  power  and  pipes  possible,  to  the  end 
that  an  adequate  selection  can  be  made. 


SOME    ECONOMICS    IN    HIGH    PRESSURE    GAS 
TRANSMISSION. 

By  EDWARD  A.  RIX. 

Mr.   President    and    Fellow    Members    of    the   Pacific 
Coast  Gas  Association. 

Last  year  I  had  the  pleasure  of  presenting  for  your 
consideration  a  paper  entitled  "The  Compression  and 
Transmission  of  Illuminating  Gas,"  in  which  the  gen- 
eral theory  of  the  subject  was  discussed,  and  methods 
shown  for  calculating  the  specific  heats  of  various  gas 
mixtures,  the  heat  developed  by  compression,  and  the 
power  required.  Also  various  losses  in  pressure  and 
power  in  pushing  gas  through  pipes,  and  some  co- 
efficients for  all  this  data,  so  that  one  could  approxi- 
mate, with  some  degree  of  certainty,  the  various  ele- 
ments of  a  practical  plant.  The  length  of  the  paper 
did  not  permit  of  bringing  it  to  such  a  closing  that 
those  not  caring  for  the  theoretical  part  could  readily 
solve  some  everyday  problems  pertaining  to  a  high- 
pressure  gas  transmission.  It  is,  therefore,  the  intent 
of  this  paper  to  briefly  supply  this  deficiency,  and  at 
the  same  time  to  introduce  the  element  of  cost  of 
compressing  gas  and  a  method  for  determining  the 
proper  size  of  pipes  so  that  the  gas  engineer  may  be 
able  to  readily  and  easily  arrive  at  the  essential  ele- 
ments in  his  problem. 

Curve  sheet  No.  20  has  been  constructed  for  this 
purpose  and  conditions  as  general  as  possible  have 
been  assumed,  and  in  order  that  it  may  be  easy  for 
anyone  to  construct  a  similar  curve  sheet  for  other 
conditions,  the  method  of  making  it  may  well  be 
explained.  Inasmuch  as  gas  is  generally  sold  and 
handled  by  the  1,000  cubic  feet,  it  seems  proper  to 
make  that  the  basis  for  quantity,  and  one  cent  per 
kilowatt  hour  seems  a  natural  base  from  which  to  cal- 
culate the  cost  of  power,  and  should  anyone  have  steam 
power  or  power  other  than  electric,  it  is  a  simple  mat- 
ter to  convert  it  to  kilowatt  hours. 


-  .  * 


s-'l  llahui 


*    =  C         „ 

*.    i^-s 

-5    !  J'j  IT,  V 


. 

>  6r  <-i 


09? 
05? 


HIGH    PRESSURE    GAS    TRANSMISSION  51 

If  a  kilowatt  hour  costs  one  cent,  a  horsepower  will 
cost  practically  three-quarters  of  a  cent,  one  horse- 
power hour=60x33,000  or  1,980,000  foot  pounds  for 
three  quarters  of  a  cent,  one  cent  would  therefore  pro- 


000 

duce  -  =2,640,000  foot  pounds. 

.  /  o 

Eight  horsepower  equals  264,000  foot  pounds,  con- 
sequently every  8  horsepower  will  cost  1-10  of  a  cent 
per  minute.  This  gives  us  the  basis  for  our  curve,  for 
if  we  lay  out  our  sheet  in  equal  divisions  of  any  size 
and  call  each  one  along  the  vertical  line  8  horsepower, 
we  can  also  make  each  division  represent  1-10  of  a 
cent,  and  each  horizontal  division  we  can  conveniently 
call  10  pounds. 

If  you  will  now  refer  to  Table  19,  the  last  table 
which  I  read  at  our  last  meeting,  it  will  be  possible 
to  construct  the  curve,  remembering  that  the  table 
is  constructed  for  100  cubic  feet  per  minute,  the  horse- 
power therein  contained  must  be  multiplied  by  10,  for 
the  1000  feet  capacity  we  are  now  considering. 

Take,  for  example,  50  pounds  gauge  pressure,  the 
brake  horse-power  required  for  100  cubic  feet  is  14.56, 
and  for  1000  would  be  145.6.  Where  the  vertical  line 
indicating  50  pounds  meets  the  horizontal  line  drawn 
from  145  horse-power  will  be  a  point  on  the  curve. 
Similarly  other  points  can  be  made,  and  joining  the 
points  together,  we  shall  have  a  cost  and  power  curve 
combined  which  will  be  very  useful  in  our  calculation. 

I  have  constructed  two  of  these  curves,  A  and  B. 
A  is  the  curve  of  single  stage  compression  and  B  for 
two-stage  compression.  Single  stage  is  rarely  used 
beyond  100  pounds  pressure,  nor  two  stage  below  90 
pounds  pressure.  You  will  note  quite  a  difference  in 
favor  of  two-stage  compression.  For  example,  at  100 
pounds  pressure  it  costs  2.35  cents  per  1000  for  two- 
stage  and  2.75  cents  for  single-stage.  In  even  a  small 
plant  using  50,000,000  feet  per  year,  the  difference 
would  be  $200  per  annum,  which  is  well  worth  saving. 

The  two-stage  curve  may  be  readily  constructed 
from  the  single-stage  curve  by  remembering  that  the 


52  SOME    ECONOMICS    IN 

intermediate  absolute  pressure  between  the  stages  is 
a  mean  proportional  between  the  initial  and  final  abso- 
lute pressure,  and,  inasmuch  as  it  takes  the  same 
power  for  each  stage,  if  we  double  the  power  required 
for  the  first  stage  we  shall  have  the  desired  results, 
thus  the  intermediate  gauge  pressure  for  200  pounds 
pressure  will  be  41  pounds.  We  note  from  the  single- 
stage  curve  that  41  pounds  requires  128  horse-power, 
consequently  twice  this  is  256  horse-power,  which,  laid 
out  on  our  curve  sheet  on  the  200-pound  vertical  line, 
will  give  us  the  point  N  on  the  two-stage  curve,  and  so 
on  for  other  points  to  complete  the  curve. 

It  must  be  understood  that  the  horse-powers  are  for 
1000  cubic  feet  per  minute,  and  the  cost  will  be  per 
1000,  and  if  you  wish  to  eliminate  the  element  of  time 
just  multiply  the  horse-power  by  33,000  and  the  re- 
sult will  be  the  foot  pounds  to  compress  1000  cubic 
feet  of  gas,  and  independent  of  time. 

If  power  costs  more  or  less  than  1  cent  per  kilowatt 
hour,  or  the  quantity  to  be  compressed  is  greater  or 
less  than  1000  cubic  feet  per  minute,  the  results  may 
be  read  from  the  curve  by  simply  using  a  correspond- 
ing proportion,  for  example : 

The  curve  shows  that  1000  cubic  feet  can  be  com- 
pressed to  20  pounds  gauge  pressure  at  the  cost  of  1 
cent,  it  follows,  therefore,  that  2000  cubic  feet  can  be 
compressed  to  20  pounds  for  2  cents,  or  if  power  costs 
2  cents  per  kilowatt  hour  instead  of  1  cent,  then  only 
one-half  the  quantity  can  be  compressed  for  1  cent,  or 
double  the  quantity  if  power  costs  but  y2  a  cent  a 
kilowatt  hour.  This  method  of  proportion,  however, 
does  not  apply  to  the  matter  of  pressure,  for  you  will 
note  that  while  a  cost  of  1  cent  gives  20  pounds  pres- 
sure, a  cost  of  2  cents  gives  58  pounds  pressure,  and  a 
cost  of  y<2  a  cent  gives  only  8  pounds  pressure.  In 
other  words,  it  costs  just  as  much  to  compress  gas 
from  0  to  8  pounds  as  it  does  from  8  to  20  pounds, 
and  just  as  much  to  compress  from  0  to  20  pounds  as 
from  20  to  58  pounds.  It  would  be  well  right  here 


HIGH    PRESSURE    GAS    TRANSMISSION  53 

to  consider  this  fact,  for  it  has  a  great  bearing  on  high- 
pressure  transmission. 

If  it  was  found,  for  example,  that  it  was  costing  1 
cent  per  1000  to  deliver  gas  through  a  certain  pipe 
at  20  pounds  pressure,  and  it  became  necessary  to 
double  the  pressure  in  order  to  supply  an  increased  de- 
mand, the  gas  company  might  consider  it  inadvisable 
because  it  might  double  the  cost.  Consulting  the  curve, 
it  will  be  seen  that  the  cost  for  compressing  at  40 
pounds  pressure  is  only  1.6  cents  per  1000  cubic  feet 
instead  of  2  cents,  as  may  be  imagined,  and  this  fact 
might  justify  the  increased  pressure,  and  the  higher 
the  pressure  the  more  the  seeming  disproportion. 

From  the  curve  take  a  geometrical  progression  of 
gauge  pressure,  5-10-20-40-80-160-320,  and  we  note  the 
corresponding  costs  of  compression  for  1000  cubic  feet 
to  be,  in  cents,  .3-.575-1.00-1.6-2.4-3.-3.9,  in  other  words, 
while  the  pressure  from  5  to  320  has  increased  sixty- 
four  times,  the  cost  of  compression  has  increased  but 
thirteen  times. 

It  must  not,  however,  be  hastily  inferred  that  be- 
cause of  this  decreasing  power  ratio  that  it  is  econom- 
ical to  compress  at  high  pressure,  because  it  may  not 
be  so  and  depends  upon  the  amount  of  gas  to  be 
pumped,  for  while  the  rate  for  1000  may  be  small  and 
make  no  material  difference  where  a  small  quantity  is 
pumped,  with  a  large  quantity  the  total  amount  of 
the  yearly  cost  of  pumping  may  exceed  so  materially 
the  interest  and  depreciation  on  a  larger  pipe  using  a 
lower  pressure  that  the  latter  installation  will  be 
deemed  preferable. 

The  question  of  whether  a  large  pipe  and  small 
pressure,  or  a  small  pipe  and  high  pressure  shall  be 
used  is  simply  a  matter  of  equating  the  relative  costs 
of  pumping,  together  with  the  interest  and  deprecia- 
tion on  the  plant,  and,  with  the  curve  given  herewith, 
it  may  be  easily  determined,  and  it  is  to  show  how  to 
make  this  determination  quite  accurately  and  simple 
that  I  have  written  this  paper.  It  has  seemed  to  me 
that  the  cost  of  attendance,  buildings,  laying  out 


54  SOME   ECONOMICS   IN 

pipe,  etc.,  for  any  one  problem  may  be  neglected,  for 
while  it  is  an  item  of  cost,  it  will  be  practically  the 
same  for  whatever  pipe  you  may  select.  We  can  then 
make  our  comparison,  using  the  market  cost  of  pipe 
and  the  cost  of  power  only,  to  which  may  be  added  the 
other  costs  after  the  size  of  the  pipe  is  determined, 
in  order  to  give  the  total  cost  per  1000  for  handling 
the  gas. 

The  power  and  cost  curves,  as  constructed,  can  be 
called  " Standard"  and  white  prints  made  from  it, 
and  upon  these  white  prints  the  pipe  curves  laid  out, 
as  will  be  shown,  and  this  same  white  print  can  be 
used  in  all  cases.  Let  us  then  take  two  examples,  one 
for  small  quantity  and  one  for  large  quantity,  and 
before  starting  at  it  let  us  make  a  general  standard 
formula,  which  will  simplify  many  of  the  calculations. 

No  plant  will  pump  less  than  10,000,000  cubic  feet 
per  year,  which  is  about  1200  per  hour,  or  at  a  less 
distance  than  10,000  feet,  consequently  take  for  a 
basis : 

10,000,000  feet  per  year=a 
10,000  feet  of  pipe=& 
1  cent  per  foot  cost  of  pipe 
10%   per  annum  interest  and  depreciation 
on  the  pipe. 

Then  equating  these  quantities  we  will  find  that  the 
pipe  cost  C  for  1000  cubic  feet  of  gas  will  be  1/10  of  a 
cent. 

For  any  other  quantity  Q,  and  length  of  pipe  L,  and 
price  of  pipe  P,  we  shall  have: 

Pipe  cost  per  1000  C  =  ^  X  ^ 

Example  1—50,000,000  cubic  feet  per  year,  or  6000 
per  hour,  50,000  feet  of  pipe,  and  power  to  cost  %  cent 
per  kilowatt  hour,  substituting  in  our  formula 

p     L  a       P  50, 000  X  10, 000, 000      _P  =  =  P 

bQ      10 W  10,000  X  50,000,000      10      10 

That  is  to  say  that  whatever  size  pipe  we  select  the 

pipe  cost  per  1000  cubic  feet  of  gas  will  be  1/10  the 

market  cost  of  pipe  per  foot. 


HIGH   PRESSURE   GAS   TRANSMISSION  55 

Taking  the  market  prices  of  to-day,  then  if  we  should 


iy^  pipe,  cost  per  1000  cubic  feet,  equals  .........  559 

1%  pipe,  cost  per  1000  cubic  feet,  equals  .........  67 

2  pipe,  cost  per  1000  cubic  feet,  equals  .........  894 

2y2  pipe,  cost  per  1000  cubic  feet,  equals  ........  1.429 

3  pipe,  cost  per  1000  cubic  feet,  equals  ........  1.875 

Having  thus  blocked  out  the  matter  of  pipe,  we  must 

find  what  pressure  it  is  necessary  to  use  to  pump  the 
gas  through  these  various  sized  pipes,  assuming  al- 
ways that  the  terminal  pressure  shall  be  one  pound 
gauge. 

You  will  remember  that  we  developed  a  formula  in 
my  paper,  read  last  year,  which  may  be  used  here  with 
accurac. 


p  ,__p,  =1d 

is  the  difference  between  the  squares  of  the  initial  and 
final  absolute  pressures. 

Q  is  the  quantity  of  gas  in  cubic  feet  per  minute. 

L  is  the  length  of  pipe,  in  feet. 

D  is  the  pipe  diameter,  in  inches. 

Substituing  in  this  equation  the  elements  in  our  prob- 
lem, we  have  : 

P  ,     P  ,  _  35X100X100X50,000  n2_175,000 

Fl  100,000  Xd  Fl  d5 

Then  Px2  XP,2  equals  for 
1^4  pipe    .......................  57,370 

1V2  pipe  .......................  23,000 

2  pipe  .......................   5,500 

2i/2  pipe  .......................   1,800 

3  pipe  .......................      700 

P2being  our  final  pressure  lib.  or  15.7  Ibs.  absolute  makes 
P22=246,  and  remembering  that  P!  2—  P22=(Pl  2—  Pl  2  + 

Pi'2 
we  have 

ikj.  pine   P./2  absolute=57,616  then  Pl  gauge=225  Ibs. 

11/2    "        "          "      =23,246     "     "  "    =140  " 

2  "       "         "      =  5,746     "     "  "    =  61  " 
2V2   "       "         "      -  2,046     "     "  "    =  31  " 

3  ^  =      946     "     "       "    =  16    " 
Now  we  are  ready  to  put  all  this  on  our  curve  sheet 

in  order  that  we  may  have  a  graphic  representation  of 


56  SOME   ECONOMICS   IN 

the  situation.  The  power  cost  curve  is  on  a  basis  of  1 
cent  per  kilowatt  hour;  if,  therefore,  we  plot  any 
other  costs  on  this  standard  sheet,  they  must  be  in- 
creased or  diminished  by  the  ratio  the  actual  power  cost 
bears  to  the  standard  power  cost  of  1  cent  per  kilo- 
watt. Our  problem  calls  for  a  power  cost  of  y2  cent 
per  kilowatt  hour,  consequently  we  must  plot  in  our 
pipe  costs  at  double  their  real  amount,  for  the  standard 
power  cost  curve  is  double  the  cost  stated  in  our  prob- 
lem. Take,  then,  the  2-inch  pipe.  We  have  found  the 
pipe  cost  to  be  .894,  which,  multiplied  by  2,  equals 
1.798,  and  the  initial  pressure  required  is  61  pounds. 
If  then  we  lay  off  the  point  P  on  the  61  pounds  pres- 
sure ordinate  and  opposite  the  cost  Line  1.798  or  1.8,  we 
shall  have  the  two-inch  pipe  established,  and  similarly 
we  can  establish  the  other  points,  and  joining  them  by 
lines  we  have  a  pipe  cost  curve  S.  C.,  which  shows 
the  cost  of  pipe  per  1000  cubic  feet  of  gas  for  all  sizes 
from  1*4  to  3  inch.  It  is  evident  that  the  total  cost 
of  pumping  the  gas  is  the  pipe  cost  plus  the  power  cost, 
consequently  if  we  add  the  pipe  cost  curve  to  the  power 
cost  curve,  we  shall  have  a  curve  of  total  cost.  Take 
the  2-inch  pipe  once  more,  and  with  dividers  measure 
off  the  distance  W.  P.,  the  cost  of  the  pipe  line,  and 
add  it  to  the  line  W.  Y.,  which  is  the  power  cost,  and  we 
have  the  point  T.,  as  the  result  of  the  addition.  Do  the 
same  for  the  other  sizes  of  pipe,  join  these  points  by 
lines,  and  we  have  the  curve  D.  F.  as  the  final  result, 
showing  the  combined  power  and  pipe  cost  per  1000 
cubic  feet  of  gas  for  all  the  sizes  of  pipe  under  consid- 
eration. 

It  is  evident  at  a  glance  that  the  2-inch  pipe  shows 
the  least  cost,  consequently  the  solution  of  our  prob- 
lem is  2-inch  pipe  at  60-pound  initial  pressure.  Total 
cost  per  1000  3.6  cents  on  a  basis  of  1  cent  per  kilowatt 
hour,  but  as  our  power  costs  !/o  a  cent  per  kilowatt 
hour,  the  total  cost  is  1.8  cents  per  1000  cubic  feet  of 
gas. 

The  graphical  method  is  very  satisfactory,  for  one 
can  see  at  a  glance  the  relations  between  the  various 


HIGH    PRESSURE    GAS    TRANSMISSION  57 

elements,  for  example :  You  will  note  that  it  cost  prac- 
tically the  same  to  pump  this  gas  through  a  l^-inch 
pipe  at  225  pounds  pressure  or  a  li/o-inch  pipe  at  140 
pounds  pressure  or  the  3 -inch  pipe  at  16  pounds  pres- 
sure, which  is  interesting. 

PROBLEM    TWO. 

Take  5000  cubic  feet  per  minute,  300,000  per  hour  or 
2,500,000,000  per  year,  through  the  same  length  pipe 
as  in  our  former  problem,  and  at  the  same  power  cost 
per  kilowatt  hour.  The  pipe  being  the  same  length 
and  the  quantity  fifty  times  greater.  We  will  have 

P  P 

— —  for  pipe  cost  per  1000  where  we  had       formerly, 

and  taking  pipe  casing  prices  up  to  12  inches,  which 
was  the  largest  I  could  get,  and  assuming  them  above 
that  size  simply  for  illustration,  we  find  the  following 
prices  per  1000  feet  of  gas  for  these  pipes : 

10  inch  equals    .25  cent  per  1000  cubic  feet  of  gas. 

12  inch  equals    .30  cent  per  1000  cubic  feet  of  gas. 

14  inch  equals    .40  cent  per  1000  cubic  feet  of  gas. 

16  inch  equals    .60  cent  per  1000  cubic  feet  of  gas. 

18  inch  equals    .80  cent  per  1000  cubic  feet  of  gas. 

20  inch  equals  1.00  cent  per  1000  cubic  feet  of  gas. 
and  the  respective  pressures  necessary  to  force  the  gas 
through  these  pipes  to  be: 

10  inch  equals  53  pounds.  16  inch  equals  12  pounds. 

12  inch  equals  26  pounds.  18  inch  equals    8  pounds. 

14  inch  equals  20  pounds.  20  inch  equals    4  pounds. 

Multiply  the  pipe  cost  by  2  and  transferring  those 
quantities  to  our  curve  sheet,  precisely  as  we  did  in  the 
other  problem,  we  have  a  pipe  curve,  E,  showing  the 
pipe  cost  per  1000  cubic  feet,  and  adding  this  to  the 
power  curve,  we  have  the  final  result  in  the  curve  F, 
which  shows  this  rather  surprising  fact,  that  the  12- 
inch  pipe  is  the  best,  and  that  it  will  carry  this  gas  at 
26  pounds  pressure,  and  at  a  cost  of  8/10  of  a  cent  per 
1000  for  power  and  pipe. 

The  curve  shows  also  that  the  gas  can  be  put  through 
the  10-inch  pipe  at  53  pounds  pressure  at  the  same 


58  SOME   ECONOMICS   IN 

cost  as  through  the  20-inch  pipe  at  4  pounds  pressure. 
I  believe  this  graphical  method  will  be  of  service  to 
you,  particularly  as  these  curve  sheets  can  be  filed 
away  and  always  used  for  quick  reference,  for  the  eye 
can  take  in  the  whole  relative  situation  at  a  glance.  I 
believe  this  method  of  handling  the  subject  removes  it 
from  the  realm  of  speculation,  and  makes  an  orderly 
comparison  of  powrer  and  pipes  possible,  to  the  end 
that  an  adequate  selection  can  be  made. 


